Question 1: Expand / Simplify the following:

$\displaystyle \text{(i) } (3x+4y)^2$

$\displaystyle (3x+4y)^2 = (3x)^2+2 \times 3x \times 4y + (4y)^2 = 9x^2 + 24xy + 16y^2$

$\displaystyle \\$

$\displaystyle \text{(ii) } (\sqrt{2}x-3y)^2$

$\displaystyle (\sqrt{2}x-3y)^2 = (\sqrt{2}x)^2 + 2 \times (\sqrt{2}x) \times (-3y) + (-3y)^2 = 2x^2 - 6\sqrt{2}xy + 9y^2$

$\displaystyle \\$

$\displaystyle \text{(iii) } \Big( 2x- \frac{1}{3x} \Big)^2$

$\displaystyle \Big( 2x- \frac{1}{3x} \Big)^2 = (2x)^2 + 2 \times (2x) \times ( - \frac{1}{3x} ) + ( \frac{1}{3x})^2 = 4x^2- \frac{4}{3} + \frac{1}{9x^2}$

$\displaystyle \\$

$\displaystyle \text{(iv) } (x-1)(x+1)(x^2+1)(x^4+1)$

$\displaystyle (x-1)(x+1)(x^2+1)(x^4+1)$

$\displaystyle = (x^2-1)(x^2+1)(x^4+1)$

$\displaystyle = (x^4-1)(x^4+1)$

$\displaystyle = (x^8-1)$

$\displaystyle \\$

$\displaystyle \text{(v) } \Big(x- \frac{1}{x} \Big) \Big(x+ \frac{1}{x} \Big) \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)$

$\displaystyle \Big(x- \frac{1}{x} \Big) \Big(x+ \frac{1}{x} \Big) \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)$

$\displaystyle = \Big(x^2- \frac{1}{x^2} \Big) \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)$

$\displaystyle = \Big(x^4- \frac{1}{x^4} \Big) \Big( x^4+ \frac{1}{x^4} \Big)$

$\displaystyle = \Big(x^8- \frac{1}{x^8} \Big)$

$\displaystyle \\$

$\displaystyle \text{(vi) } (2x- \frac{3}{x} +1)(2x+ \frac{3}{x} + 1)$

$\displaystyle (2x- \frac{3}{x} +1)(2x+ \frac{3}{x} + 1)$

$\displaystyle = (2x+1 - \frac{3}{x} )(2x+1 + \frac{3}{x} )$

$\displaystyle = (2x+1)^2 - \Big( \frac{3}{x} \Big)^2$

$\displaystyle = 4x^2 + 4x + 1 - \frac{9}{x^2}$

$\displaystyle \\$

(vii) $\displaystyle (2x+5y+3)(2x+5y+4)$

Let $\displaystyle 2x+5 = a$

$\displaystyle \Rightarrow (2x+5y+3)(2x+5y+4)$

$\displaystyle = (a+3)(a+4)$

$\displaystyle = a^2 + 7a + 12$

$\displaystyle = (2x+5)^2 + 7(2x+5)+ 12$

$\displaystyle = 4x^2 + 20x + 25 + 14x + 35 + 12$

$\displaystyle = 4x^2 + 34x + 72$

$\displaystyle \\$

(viii) $\displaystyle (1.5x^2-0.3y^2)(1.5x^2+0.3y^2)$

$\displaystyle (1.5x^2-0.3y^2)(1.5x^2+0.3y^2) = (1.5x^2)^2 - (0.3y^2)^2 = 2.25x^4-0.09y^4$

$\displaystyle \\$

(ix) $\displaystyle (x^3-3x^2-x)(x^2-3x+1)$

$\displaystyle (x^3-3x^2-x)(x^2-3x+1)$

$\displaystyle = x(x^2+3x+1)(x^2-3x+1)$

$\displaystyle = x \Big( (x^2+3x)^2 - 1 \Big)$

$\displaystyle = x(x^4-6x^3+9x^2-1)$

$\displaystyle = x^5-6x^4+9x^3-1$

$\displaystyle \\$

(x) $\displaystyle (2x^4-4x^2+1)(2x^4-4x^2-1)$

$\displaystyle (2x^4-4x^2+1)(2x^4-4x^2-1)$

$\displaystyle = (2x^4-4x^2)^2 - 1$

$\displaystyle = 4x^8-16x^6+16x^4-1$

$\displaystyle \\$

(xi) $\displaystyle \Big(x+ \frac{2}{x} -3\Big) \Big(x- \frac{2}{x} -3 \Big)$

$\displaystyle \Big(x+ \frac{2}{x} -3\Big) \Big(x- \frac{2}{x} -3 \Big)$

$\displaystyle = \Big( (x - 3 )+ \frac{2}{x} \Big) \Big( (x-3) - \frac{2}{x} \Big)$

$\displaystyle = x^2-6x+9 - \frac{4}{x^2}$

$\displaystyle \\$

(xii) $\displaystyle (5-2x)(5+2x)(25+4x^2)$

$\displaystyle (5-2x)(5+2x)(25+4x^2) = (25-4x^2)(25+4x^2) = 625-16x^4$

$\displaystyle \\$

(xiii) $\displaystyle (x+2y+3)(x+2y+7)$

$\displaystyle (x+2y+3)(x+2y+7)$

$\displaystyle = (x+2y)^2 +10(x+2y) + 21$

$\displaystyle = x^2+4xy+4y^2+10x+20y+21$

$\displaystyle \\$

(xiv) $\displaystyle (x+1)(x+2)(x+3)$

$\displaystyle (x+1)(x+2)(x+3)$

$\displaystyle = (x^2+3x+2)(x+3)$

$\displaystyle = x^3+3x^2+2x+3x^2+9x+6$

$\displaystyle = x^3 +6x^2 +11x +6$

$\displaystyle \\$

(xv) $\displaystyle (a+2b+c)^2$

Note: We will use the following identify: $\displaystyle {(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)$

$\displaystyle (a+2b+c)^2 = a^2 + 4b^2 +c^2 + 4ab + 4bc+ 2ca$

$\displaystyle \\$

(xvi) $\displaystyle \Big( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \Big)^2$

$\displaystyle \Big( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \Big)^2$

$\displaystyle = (x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2) - (x^4+y^4+z^4 - 2x^2y^2-2y^2z^2+2z^2x^2)$

$\displaystyle = x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2 -x^4-y^4-z^4 + 2x^2y^2+2y^2z^2-2z^2x^2$

$\displaystyle = 4x^2y^2 - 4z^2x^2$

$\displaystyle = 4x^2 (y^2-z^2)$

$\displaystyle = 4x^2(y+z)(y-z)$

$\displaystyle \\$

(xvii) $\displaystyle (-2x+3y+2z)^2$

$\displaystyle (-2x+3y+2z)^2 = 4x^2+9y^2+4z^2 -12xy+12yz-8zx$

$\displaystyle \\$

(xviii) $\displaystyle (a+b+c)^2 - (a-b+c)^2$

$\displaystyle (a+b+c)^2 - (a-b+c)^2$

$\displaystyle = \Big( a+b+c + a-b+c \Big) \Big( a+b+c -a +b -c \Big)$

$\displaystyle = ( 2a+2c ) ( 2b ) = 4ab + 4bc$

$\displaystyle \\$

(xix) $\displaystyle (x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2$

$\displaystyle (x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2$

$\displaystyle = \Big( x^2+y^2+z^2 + x^2-y^2+z^2 \Big) \Big( x^2+y^2+z^2 - x^2 +y^2 - z^2 \Big)$

$\displaystyle = (2x^2+2z^2)(2y^2) = 4x^2y^2+4y^2z^2$

$\displaystyle \\$

(xx) $\displaystyle (x+y+z)^2+ \Big( x+ \frac{y}{2} + \frac{z}{3} \Big)^2 - \Big( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} \Big)^2$

$\displaystyle (x+y+z)^2+ \Big( x+ \frac{y}{2} + \frac{z}{3} \Big)^2 - \Big( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} \Big)^2$

$\displaystyle = x^2+y^2+z^2 + 2xy + 2yz+2zx + x^2 + \frac{1}{4} y^2 + \frac{1}{9} z^2 + xy + \frac{1}{3} yz +$

$\displaystyle \frac{2}{3} zx - \frac{1}{4} x^2 - \frac{1}{9} y^2 - \frac{1}{16} z^2 - \frac{1}{3} xy - \frac{1}{6} yz - \frac{1}{4} zx$

$\displaystyle = x^2( 1 + 1 - \frac{1}{4} ) + y^2(1+ \frac{1}{4} - \frac{1}{9} ) +z^2(1 + \frac{1}{9} -\frac{1}{16} ) \\ +xy(2+1- \frac{1}{3} )$

$\displaystyle + yz(2 + \frac{1}{3} - \frac{1}{6} ) +zx(2+ \frac{2}{3} - \frac{1}{4} )$

$\displaystyle = \frac{7}{4} x^2 + \frac{41}{36} y^2 + \frac{151}{144} z^2 + \frac{8}{3} xy + \frac{13}{6} yz + \frac{29}{12} zx$

$\displaystyle \\$

(xx) $\displaystyle (x^2-x+1)^2-(x^2+x+1)^2$

$\displaystyle (x^2-x+1)^2-(x^2+x+1)^2$

$\displaystyle = (x^2-x+1 + x^2+x+1)(x^2-x+1 - x^2-x-1)$

$\displaystyle = (2x^2+2)(-2x) = -4x^3-4x$

$\displaystyle \\$

(xxi) $\displaystyle \Big( x+ \frac{2}{x} \Big)^2 + \Big( x- \frac{2}{x} \Big)^2$

We will use the identity $\displaystyle {a^3+b^3= \ (a+b)}^3-3ab(a+b)$

$\displaystyle \Rightarrow a = x+ \frac{2}{x}$

$\displaystyle \Rightarrow b = x- \frac{2}{x}$

$\displaystyle \Rightarrow a + b = 2x$

$\displaystyle \Big( x+ \frac{2}{x} \Big)^2 + \Big( x- \frac{2}{x} \Big)^2$

$\displaystyle = (2x)^3 - 3(x + \frac{2}{x} )( x- \frac{2}{x} ) (2x)$

$\displaystyle = 8x^3-6x^3+ \frac{24}{x} = 2x^3 + \frac{24}{x}$

$\displaystyle \\$

(xxii) $\displaystyle (2x-5y)^3-(2x+5y)^3$

We will use the identity $\displaystyle {a^3-b^3=\ (a-b)}^3+3ab(a-b)$

$\displaystyle \Rightarrow a = 2x-5y \ and \ b = 2x+5y$

$\displaystyle \Rightarrow a - b = 2x-5y-2x-5y = -10y$

Therefore $\displaystyle (2x-5y)^3-(2x+5y)^3$

$\displaystyle = (-10y)^3 + 3 (2x-5y)(2x+5y)(-10y)$

$\displaystyle = -1000y^3 -120yx^2+750y^3$

$\displaystyle = -250y^3 -120yx^2$

$\displaystyle \\$

(xxiii) $\displaystyle (4x-5y)(16x^2+20xy+25y^2)$

We will use the identity $\displaystyle a^3-b^3 = (a-b)(a^2 +ab + b^2)$

$\displaystyle (4x-5y)(16x^2+20xy+25y^2)$

$\displaystyle = (4x-5y)\Big[ (4x)^2 + (4x)(5y) + (5y)^2 \Big]$

$\displaystyle = (4x)^3 -(5y)^3 = 64x^3 - 125y^3$

$\displaystyle \\$

(xxiv) $\displaystyle (x^3+1)(x^6-x^3+1)$

We will use the identity $\displaystyle a^3-b^3 = (a-b)(a^2 +ab + b^2)$

$\displaystyle (x^3+1)(x^6-x^3+1)$

$\displaystyle = (x^3+1) \Big[ (x^3)^2 -(x^3)(1) +(1)^2 \Big]$

$\displaystyle = (x^3)^3 +(1)^3 = x^9 +1$

$\displaystyle \\$

(xxv) $\displaystyle (4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx)$

We will use the identity $\displaystyle a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\displaystyle (4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx)$

$\displaystyle = (4x-3y+2z) \Big[ (4x)^2 +(-3y)^2 + (2z)^2 - (4x)(-3y) - (-3y)(2z) - (2z)(4x) \Big]$

$\displaystyle = (4x)^3+(-3y)^3+(2z)^3 - 3(4x)(-3y)(2z)$

$\displaystyle = 64x^3 - 27y^3+8z^3 +72xyz$

$\displaystyle \\$

Question 2: Evaluate the following identities:

$\displaystyle \text{(i) } 103 \times 97$

$\displaystyle 103 \times 97 = (100+3)(100-3) = 100^2 - 3^2 = 10000-9 = 9991$

$\displaystyle \\$

$\displaystyle \text{(ii) } (97)^2$

$\displaystyle (97)^2 = (100-3)^2 = 10000 -600 +9 = 9409$

$\displaystyle \\$

$\displaystyle \text{(iii) } 0.54 \times 0.54 -0.46 \times 0.46$

$\displaystyle 0.54 \times 0.54 -0.46 \times 0.46 = 0.54^2-0.46^2 = (0.54+0.46)(0.54-0.46) = 1 \times 0.08 = 0.08$

$\displaystyle \\$

$\displaystyle \text{(iv) } (0.98)^2$

$\displaystyle (0.98)^2 = (1-0.02)^2 = 1 -0.04+0.0004 = 0.9604$

$\displaystyle \\$

$\displaystyle \text{(v) } 991 \times 1009$

$\displaystyle 991 \times 1009 = (1000 -9) (1000 + 9) = 1000^2 -9^2 = 1000000-81 = 999919$

$\displaystyle \\$

$\displaystyle \text{(vi) } (1002)^3$

We will use the identity $\displaystyle {(a+b)}^3=\ a^3+b^3+3ab(a+b)$

$\displaystyle (1002)^3 = (1000+2)^3 = 1000^3 + 2^3 + 3\times 1000 \times 2 (1000 + 2)$

$\displaystyle = 1000000000+8 + 6012000= 1006012008$

$\displaystyle \\$

(vii) $\displaystyle (999)^3$

We will use the identity $\displaystyle {(a-b)}^3=\ a^3-b^3-3ab(a-b)$

$\displaystyle (999)^3 = (1000 - 1)^3 = 1000^3 -1^3-3 \times 1000 \times 1 (1000-1)$

$\displaystyle = 1000000000-1-299700 = 997002999$

$\displaystyle \\$

(viii) $\displaystyle (103)^3$

We will use the identity $\displaystyle {(a+b)}^3=\ a^3+b^3+3ab(a+b)$

$\displaystyle (103)^3 = (100 + 3)^3 = 100^3 + 3^3 + 3 \times 100 \times 3 (100+3)$

$\displaystyle = 1000000+27+92700 = 1092727$

$\displaystyle \\$

(ix) $\displaystyle (10.4)^3$

$\displaystyle (10.4)^3 = (10+0.4)^3 = 10^3 +0.4^3 + 3 \times 10 \times 0.4 (10+0.4)$

$\displaystyle = 1000 + 0.064+124.8 = 1124.864$

$\displaystyle \\$

(x) $\displaystyle (99)^3$

We will use the identity $\displaystyle {(a-b)}^3=\ a^3-b^3-3ab(a-b)$

$\displaystyle (99)^3 = (100 -1 )^3 = 100^3 - 1^3 - 3 \times 100 \times 1 (100 -1 )$

$\displaystyle = 1000000-1-29700=970299$

$\displaystyle \\$

(xi) $\displaystyle 111^3 - 89^3$

We will use the identity $\displaystyle {a^3-b^3=\ (a-b)}^3+3ab(a-b)$

$\displaystyle 111^3 - 89^3 = (100+11)^3 - (100-11)^3$

$\displaystyle = (100+11-100+11)^3 + 3 (100+11)(100-11) (22)$

$\displaystyle =22^3 + 3 \times 111 \times 89 \times 22$

$\displaystyle = 662662$

$\displaystyle \\$

(xii) $\displaystyle 104^3+96^3$

We will use the identity $\displaystyle a^3+b^3= (a+b)^3-3ab(a+b)$

$\displaystyle 104^3+96^3$

$\displaystyle = (100+4)^3 + (100-4)^3$

$\displaystyle = (100+4+100-4) - \Big[ (100+4+100-4)^3 - 3 (100+4)(100-4)(100+4+100-4) \Big]$

$\displaystyle = 200^3 - 3 \times 104 \times 96 \times 200$

$\displaystyle = 2009600$

$\displaystyle \\$

(xiii) $\displaystyle \Big( \frac{1}{2} \Big)^3 + \Big( \frac{1}{3} \Big)^3 - \Big( \frac{5}{6} \Big)^3$

We are going to use this identity $\displaystyle a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\displaystyle \Big( \frac{1}{2} \Big)^3 + \Big( \frac{1}{3} \Big)^3 - \Big( \frac{5}{6} \Big)^3$

$\displaystyle = ( \frac{1}{2} + \frac{1}{3} - \frac{5}{6} ) \Big ( ( \frac{1}{2})^2 + ( \frac{1}{3} )^2 + (- \frac{5}{6} )^2 - \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{2} \Big) + 3 \times \frac{1}{2} \times \frac{1}{3} \times (- \frac{5}{6} )$

$\displaystyle = \frac{0}{6} \Big ( ( \frac{1}{2})^2 +( \frac{1}{3} )^2 + (- \frac{5}{6})^2 - \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{2} \Big) - \frac{5}{12}$

$\displaystyle = \frac{-5}{12}$

$\displaystyle \\$

Question 3:

(i) If the number $\displaystyle a$ is $\displaystyle 7$ more than number $\displaystyle b$ and the sum of the squares of $\displaystyle a$ and $\displaystyle b$ is $\displaystyle 85$, find the product of $\displaystyle ab$

$\displaystyle a+b = 7 \Rightarrow a - b = 7$

$\displaystyle a^2 + b^2 = 85$

Therefore $\displaystyle (a-b)^2 + 2ab = 85$

$\displaystyle \Rightarrow 7^2 + 2ab = 85$

$\displaystyle \Rightarrow 2ab = 85 - 49 = 36$

$\displaystyle \Rightarrow ab = 18$

$\displaystyle \\$

(ii) If the number $\displaystyle x$ is $\displaystyle 3$ less than the number $\displaystyle y$ and the sum if the square of $\displaystyle x$ and $\displaystyle y$ is $\displaystyle 29$, find $\displaystyle xy$

$\displaystyle x+3 = 7 \Rightarrow x - y = -3$

$\displaystyle x^2 + y^2 = 29$

$\displaystyle (x-y)^2 +2xy = 29$

$\displaystyle \Rightarrow (-3)^2 + 2xy = 29$

$\displaystyle \Rightarrow 2xy = 29 - 9 = 20$

$\displaystyle \Rightarrow xy = 10$

$\displaystyle \\$

(iii) If the sum of two numbers is $\displaystyle 7$ and the sum of their cubes is $\displaystyle 133$, find the sum of their squares.

$\displaystyle a+ b = 7$

$\displaystyle a^3 + b^3 = 133$

$\displaystyle (a+b)^3 = a^3 +b^3 +3ab(a+b)$

$\displaystyle 7^2 = 133 + 3ab (7)$

$\displaystyle 21ab = 133 - 49 = 84$

$\displaystyle \Rightarrow ab = 4$

Therefore $\displaystyle (a+b)^2 -2ab = a^2 + b^2$

$\displaystyle \Rightarrow 7^2 - 8 = a^2 +b^2$

$\displaystyle \Rightarrow a^2 +b^2 = 41$

$\displaystyle \\$

Question 4:

(i) If $\displaystyle x + \frac{1}{x} = 6$, find (a) $\displaystyle x^2 + \frac{1}{x^2}$ (b) $\displaystyle x^4 + \frac{1}{x^4}$

(a) $\displaystyle x + \frac{1}{x} = 6$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 36$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} + 2 = 36$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 34$

(b) $\displaystyle x^2 + \frac{1}{x^2} = 34$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 34^2$

$\displaystyle \Rightarrow x^4 + \frac{1}{x^4} + 2 = 1156$

$\displaystyle \Rightarrow x^4 + \frac{1}{x^4} = 1154$

$\displaystyle \\$

(ii) If $\displaystyle x = \frac{1}{x - 2\sqrt{3}}$, find (a) $\displaystyle x- \frac{1}{x}$ (b) $\displaystyle x- \frac{1}{x}$ (c) $\displaystyle x^2- \frac{1}{x^2}$

(a) $\displaystyle x = \frac{1}{x - 2\sqrt{3}}$

$\displaystyle x(x - 2\sqrt{3}) = 1$

$\displaystyle x^2 - 1 = 2\sqrt{3}x$

$\displaystyle \frac{x^2-1}{x} = 2\sqrt{3}$

$\displaystyle x- \frac{1}{x} = 2\sqrt{3}$

(b) Since $\displaystyle (a+b)^2 - (a-b)^2 = 4ab$

Therefore $\displaystyle \Big( x + \frac{1}{x} \Big)^2 - \Big( x - \frac{1}{x} \Big)^2 = 4$

$\displaystyle \Rightarrow\Big( x + \frac{1}{x} - \Big)^2 - (2\sqrt{3})^2 = 4$

$\displaystyle \Rightarrow\Big( x + \frac{1}{x} - \Big)^2 = 4 + (2\sqrt{3})^2$

$\displaystyle \Rightarrow\Big( x + \frac{1}{x} - \Big)^2 =16$

$\displaystyle \Rightarrow x + \frac{1}{x} - = \pm 4$

(c) $\displaystyle x- \frac{1}{x} = 2\sqrt{3}$

$\displaystyle \Rightarrow x + \frac{1}{x} - = \pm 4$

Therefore $\displaystyle (x- \frac{1}{x} ) (x + \frac{1}{x} ) = 2\sqrt{3} \times \pm 4 = \pm 8\sqrt{3}$

$\displaystyle \\$

(iii) If $\displaystyle x^2+ \frac{1}{x^2} = 27$, find (a) $\displaystyle x+ \frac{1}{x}$ (b) $\displaystyle x- \frac{1}{x}$

(a) $\displaystyle x^2+ \frac{1}{x^2} = 27$

$\displaystyle \Big( x + \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} +2$

$\displaystyle \Big( x + \frac{1}{x} \Big)^2 = 27 +2$

$\displaystyle \Big( x + \frac{1}{x} \Big) = \pm \sqrt{29}$

(b) $\displaystyle x^2+ \frac{1}{x^2} = 27$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = 27 - 2$

$\displaystyle \Big( x - \frac{1}{x} \Big) = \pm 5$

$\displaystyle \\$

$\displaystyle \text{(iv) } x + \frac{1}{x} = 11$, find the value of $\displaystyle x^2 + \frac{1}{x^2}$

$\displaystyle x + \frac{1}{x} = 11$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 121$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} + 2 = 121$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 119$

$\displaystyle \\$

$\displaystyle \text{(v) } x - \frac{1}{x} = -1$, find the value of $\displaystyle x^2 + \frac{1}{x^2}$

$\displaystyle x - \frac{1}{x} = -1$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 = 1$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} - 2 = 1$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 3$

$\displaystyle \\$

$\displaystyle \text{(vi) } x + \frac{1}{x} = \sqrt{5}$, find the value of $\displaystyle x^2 + \frac{1}{x^2}$ and $\displaystyle x^4 + \frac{1}{x^4}$

$\displaystyle x + \frac{1}{x} = \sqrt{5}$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 5$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} + 2 = 5$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 3$

Now

$\displaystyle x^2 + \frac{1}{x^2} = 3$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 9$

$\displaystyle \Rightarrow x^4 + \frac{1}{x^4} + 2 = 9$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 7$

$\displaystyle \\$

(vii) If $\displaystyle x^2 + \frac{1}{x^2} = 66$, find the value of $\displaystyle x- \frac{1}{x}$

$\displaystyle x^2+ \frac{1}{x^2} = 66$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = 66 - 2$

$\displaystyle \Big( x - \frac{1}{x} \Big) = \pm 8$

$\displaystyle \\$

(viii) If $\displaystyle x^2 + \frac{1}{x^2} = 79$, find the value of $\displaystyle x+ \frac{1}{x}$

$\displaystyle x^2+ \frac{1}{x^2} = 79$

$\displaystyle \Big( x + \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} +2$

$\displaystyle \Big( x + \frac{1}{x} \Big)^2 = 79 +2$

$\displaystyle \Big( x + \frac{1}{x} \Big) = \pm 9$

$\displaystyle \\$

(ix) If $\displaystyle a^2-3a-1=0$, find the value of $\displaystyle a^2 + \frac{1}{a^2}$

$\displaystyle a^2-3a-1=0$

$\displaystyle \Rightarrow a^2 -1 = 3a$

$\displaystyle \Rightarrow \frac{a^2-1}{a} = 3$

$\displaystyle \Rightarrow a - \frac{1}{a} = 3$

$\displaystyle \Big( a - \frac{1}{a} \Big)^2 = a^2 + \frac{1}{a^2} -2$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = 9 - 2$

$\displaystyle \Big( x - \frac{1}{x} \Big) = \pm 7$

$\displaystyle \\$

(x) If $\displaystyle x^2 + \frac{1}{x^2} = 7$, find the value of $\displaystyle 3x^2- \frac{3}{x^2}$

$\displaystyle x^2+ \frac{1}{x^2} = 7$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = x^2 + \frac{1}{x^2} -2$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = 7 - 2$

$\displaystyle \Big( x - \frac{1}{x} \Big) = \pm \sqrt{5}$

Now $\displaystyle 3x^2- \frac{3}{x^2} = 3 ( x^2- \frac{1}{x^2} ) = \pm 3\sqrt{5}$

$\displaystyle \\$

(xi) If $\displaystyle a = \frac{1}{a-5}$, find the value of (a) $\displaystyle a - \frac{1}{a}$ (b) $\displaystyle a + \frac{1}{a}$ (c) $\displaystyle a^2 + \frac{1}{a^2}$

(a) $\displaystyle a = \frac{1}{a-5}$

$\displaystyle a^2 - 5a = 1$

$\displaystyle \Rightarrow a^2 - 1 = 5a$

$\displaystyle \Rightarrow a - \frac{1}{a} = 5$

(b) $\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^2 = \Rightarrow \Big( a - \frac{1}{a} \Big)^2 +4$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 29$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big) = \pm \sqrt{29}$

(c) $\displaystyle a + \frac{1}{a} = \pm \sqrt{29}$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 29$

$\displaystyle \Rightarrow a^2 + \frac{1}{a^2} + 2 = 29$

$\displaystyle \Rightarrow a^2 + \frac{1}{a^2} = 27$

$\displaystyle \\$

(xii) If $\displaystyle a^2-3a+1 = 0$, find the values of (a) $\displaystyle a^2 + \frac{1}{a^2}$ (b) $\displaystyle a^3 + \frac{1}{a^3}$

$\displaystyle a^2-3a+1 = 0$

$\displaystyle a^2+1 = 3a$

$\displaystyle \Rightarrow a^2 + 1 = 3a$

$\displaystyle \Rightarrow a + \frac{1}{a} = 3$

(a) $\displaystyle a + \frac{1}{a} = 3$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^2 = 9$

$\displaystyle \Rightarrow a^2 + \frac{1}{a^2} + 2 = 9$

$\displaystyle \Rightarrow a^2 + \frac{1}{a^2} = 7$

(b) $\displaystyle a + \frac{1}{a} = 3$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^3 = a^3 + \frac{1}{a^3} + 3. a . \frac{1}{a} (a + \frac{1}{a} )$

$\displaystyle \Rightarrow \Big( a + \frac{1}{a} \Big)^3 = a^3 + \frac{1}{a^3} + 3 (a + \frac{1}{a} )$

$\displaystyle a^3 + \frac{1}{a^3} = 3^3 -3(3) = 27 -9 = 18$

$\displaystyle \\$

(xiii) If $\displaystyle x = 5 - 2\sqrt{6}$ find the value of $\displaystyle \sqrt{x}+ \frac{1}{\sqrt{x}}$

$\displaystyle x = 5 - 2\sqrt{6}$

$\displaystyle \Big( \sqrt{x} + \frac{1}{\sqrt{x}} \Big)^2 = x + \frac{1}{x} + 2$

$\displaystyle = 5 - 2\sqrt{6} + \frac{1}{5 - 2\sqrt{6}} + 2$

$\displaystyle = \frac{25+24-20\sqrt{6}+1+10-4\sqrt{6}}{5 - 2\sqrt{6}}$

$\displaystyle = \frac{60-24\sqrt{6}}{5 - 2\sqrt{6}}$

$\displaystyle = \frac{12(5 - 2\sqrt{6})}{5 - 2\sqrt{6}} = 12$

Therefore $\displaystyle \sqrt{x}+ \frac{1}{\sqrt{x}} = \pm \sqrt{12}$

$\displaystyle \\$

(xiv) If $\displaystyle a+b+c = 0$ and $\displaystyle a^2+b^2+c^2=16$, find the value of $\displaystyle ab+bc+ca$

Given: $\displaystyle a+b+c = 0$ and $\displaystyle a^2+b^2+c^2=16$

We know: $\displaystyle (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$\displaystyle \Rightarrow ab+bc+ca = \frac{0-16}{2} = -8$

$\displaystyle \\$

(xv) If $\displaystyle a^2+b^2+c^2=16$ and $\displaystyle ab+bc+ca=10$, find the value of $\displaystyle a+b+c$

Given: $\displaystyle a^2+b^2+c^2=16$ and $\displaystyle ab+bc+ca=10$

We know: $\displaystyle (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$\displaystyle \Rightarrow (a+b+c)^2 = 16+2 \times 10 = 36$

$\displaystyle \Rightarrow a+b+c = \pm 6$

$\displaystyle \\$

(xvi) If $\displaystyle a+b+c = 9$ and $\displaystyle ab+bc+ca=23$, find the value of $\displaystyle a^2+b^2+c^2$

Given: $\displaystyle a+b+c = 9$ and $\displaystyle ab+bc+ca=23$

We know: $\displaystyle (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$\displaystyle \Rightarrow a^2 + b^2 + c^2 = 81-46 = 35$

$\displaystyle \\$

(xvii) If $\displaystyle x+y-z=4$ and $\displaystyle x^2+y^2+z^2=38$, find the value of $\displaystyle xy-yz-zx$

Given: $\displaystyle x+y-z=4$ and $\displaystyle x^2+y^2+z^2=38$

We know: $\displaystyle (x+y-z)^2 = x^2 + y^2 + z^2 + 2(xy - yz - zx)$

$\displaystyle \Rightarrow xy - yz - zx = \frac{16-38}{2} = -11$

$\displaystyle \\$

(xviii) If $\displaystyle x^2 + \frac{1}{x^2} = 7$, find the value of $\displaystyle x^3+ \frac{1}{x^3}$

Given $\displaystyle x^2 + \frac{1}{x^2} = 7$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =9$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big) = \pm 3$

If $\displaystyle x + \frac{1}{x} = 3$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 3^3 -3(3) = 18$

Similarly, if $\displaystyle x + \frac{1}{x} = -3$

Then $\displaystyle x^3 + \frac{1}{x^3} = (-3)^3 -3(-3) = -18$

$\displaystyle \\$

(xix) If $\displaystyle x^2 + \frac{1}{x^2} = 83$, find the value of $\displaystyle x^3- \frac{1}{x^3}$

Given $\displaystyle x^2 + \frac{1}{x^2} = 83$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =81$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big) = \pm 9$

If $\displaystyle x - \frac{1}{x} = 9$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )$

$\displaystyle x^3 - \frac{1}{x^3} = 9^3 +3(9) = 756$

Similarly, if $\displaystyle x - \frac{1}{x} = -9$

Then $\displaystyle x^3 - \frac{1}{x^3} = (-9)^3 +3(-9) = -756$

$\displaystyle \\$

(xx) If $\displaystyle x^4 + \frac{1}{x^4} = 47$, find the value of $\displaystyle x^3+ \frac{1}{x^3}$

Given $\displaystyle x^4 + \frac{1}{x^4} = 47$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =49$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) = 7$

Now $\displaystyle \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =9$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big) = \pm 3$

If $\displaystyle x + \frac{1}{x} = 3$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 3^3 -3(3) = 18$

Similarly, if $\displaystyle x + \frac{1}{x} = -3$

Then $\displaystyle x^3 + \frac{1}{x^3} = (-3)^3 -3(-3) = -18$

$\displaystyle \\$

(xxi) If $\displaystyle a+b = 10$ and $\displaystyle ab = 21$, find the value of $\displaystyle a^3+b^3$

Given $\displaystyle a+b = 10$ and $\displaystyle ab = 21$

$\displaystyle (a+b)^3 = a^3 + b^3 +3ab(a+b)$

$\displaystyle \Rightarrow a^3 + b^3 = 10^3 - 3 \times 21 \times 10 = 1000-630 = 370$

$\displaystyle \\$

(xxii) If $\displaystyle a-b = 4$ and $\displaystyle ab = 21$, find the value of $\displaystyle a^3-b^3$

Given $\displaystyle a-b = 4$ and $\displaystyle ab = 21$

$\displaystyle (a-b)^3 = a^3 - b^3 - 3ab(a-b)$

$\displaystyle \Rightarrow a^3 - b^3 = 4^3 + 3 \times 21 \times 4 = 64+252 = 316$

$\displaystyle \\$

(xxiii) $\displaystyle x + \frac{1}{x} = 5$, find $\displaystyle x^3 + \frac{1}{x^3}$

If $\displaystyle x + \frac{1}{x} = 5$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 5^3 -3(5) = 110$

$\displaystyle \\$

(xxiv) $\displaystyle x - \frac{1}{x} = 7$, find $\displaystyle x^3 - \frac{1}{x^3}$

If $\displaystyle x - \frac{1}{x} = 7$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )$

$\displaystyle x^3 - \frac{1}{x^3} = 7^3 + 3(7) = 364$

$\displaystyle \\$

(xxv) $\displaystyle x^2 + \frac{1}{x^2} = 51$, find $\displaystyle x^3 - \frac{1}{x^3}$

Given $\displaystyle x^2 + \frac{1}{x^2} = 51$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =49$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big) = \pm 7$

If $\displaystyle x - \frac{1}{x} = 7$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )$

$\displaystyle x^3 - \frac{1}{x^3} = 7^3 + 3(7) = 364$

Similarly, if $\displaystyle x - \frac{1}{x} = -7$

Then $\displaystyle x^3 - \frac{1}{x^3} = (-7)^3 +3(-7) = -364$

$\displaystyle \\$

(xxvi) $\displaystyle x^2 + \frac{1}{x^2} = 98$, find $\displaystyle x^3 + \frac{1}{x^3}$

Given $\displaystyle x^2 + \frac{1}{x^2} = 98$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =100$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big) = 10$

Now $\displaystyle x + \frac{1}{x} = \pm 10$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 10^3 -3(10) = 100-30 = 970$

Similarly, if $\displaystyle x + \frac{1}{x} = -10$

Then $\displaystyle x^3 + \frac{1}{x^3} = (-10)^3 -3(-10) = -970$

$\displaystyle \\$

(xxvii) $\displaystyle x^4 + \frac{1}{x^4} = 119$, find $\displaystyle x^3 - \frac{1}{x^3}$

Given $\displaystyle x^4 + \frac{1}{x^4} = 119$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =121$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) = 11$

Given $\displaystyle x^2 + \frac{1}{x^2} = 11$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} - 2$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^2 =9$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big) = \pm 3$

If $\displaystyle x - \frac{1}{x} = 3$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )$

$\displaystyle x^3 - \frac{1}{x^3} = 3^3 +3(3) = 36$

Similarly, if $\displaystyle x - \frac{1}{x} = -3$

Then $\displaystyle x^3 - \frac{1}{x^3} = (-3)^3 +3(-3) = -36$

$\displaystyle \\$

(xxviii) $\displaystyle x + \frac{1}{x} = 5$, find $\displaystyle x^2 + \frac{1}{x^2}$, $\displaystyle x^3 + \frac{1}{x^3}$ and $\displaystyle x^4 + \frac{1}{x^4}$

$\displaystyle x + \frac{1}{x} = 5$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 = 25$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} + 2 = 25$

$\displaystyle \Rightarrow x^2 + \frac{1}{x^2} = 23$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 5^3 -3(5) = 110$

Now $\displaystyle x^2 + \frac{1}{x^2} = 23$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 = 529$

$\displaystyle \Rightarrow x^4 + \frac{1}{x^4} + 2 = 529$

$\displaystyle \Rightarrow x^4 + \frac{1}{x^4} = 527$

$\displaystyle \\$

(xxix) $\displaystyle x^4 + \frac{1}{x^4} = 194$, find $\displaystyle x^3 + \frac{1}{x^3}$, $\displaystyle x^2 + \frac{1}{x^2}$ and $\displaystyle x + \frac{1}{x}$

Given $\displaystyle x^4 + \frac{1}{x^4} = 194$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =x^4 + \frac{1}{x^4} + 2$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big)^2 =196$

$\displaystyle \Rightarrow \Big( x^2 + \frac{1}{x^2} \Big) = 14$

Therefore $\displaystyle \Big( x + \frac{1}{x} \Big)^2 =x^2 + \frac{1}{x^2} + 2$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =16$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big) = \pm 4$

If $\displaystyle x + \frac{1}{x} = 4$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52$

Similarly, if $\displaystyle x + \frac{1}{x} = -4$

Then $\displaystyle x^3 + \frac{1}{x^3} = (-4)^3 +3(-4) = -52$

$\displaystyle \\$

(xxx) $\displaystyle x - \frac{1}{x} = 3+2\sqrt{2}$, find $\displaystyle x^3 - \frac{1}{x^3}$

$\displaystyle x - \frac{1}{x} = 3+2\sqrt{2}$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3. x . \frac{1}{x} (x - \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x - \frac{1}{x} \Big)^3 = x^3 - \frac{1}{x^3} - 3 (x - \frac{1}{x} )$

$\displaystyle x^3 - \frac{1}{x^3} = (3+2\sqrt{2})^3 +3(3+2\sqrt{2})$

$\displaystyle = (3+2\sqrt{2}) \Big[ (3+2\sqrt{2})^2 +3 \Big]$

$\displaystyle = (3+2\sqrt{2})(20+12\sqrt{2}) = 108+76\sqrt{2}$

$\displaystyle \\$

(xxxi) $\displaystyle x = \frac{1}{4-x}$, find $\displaystyle x + \frac{1}{x}$, $\displaystyle x^3 + \frac{1}{x^3}$ and $\displaystyle x^6 + \frac{1}{x^6}$

Given $\displaystyle x = \frac{1}{4-x}$

$\displaystyle \Rightarrow x^2+1=4x \Rightarrow x + \frac{1}{x} = 4$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52$

$\displaystyle \Rightarrow \Big( x^3 + \frac{1}{x^3} \Big)^2 = 52$

$\displaystyle \Rightarrow x^6 + \frac{1}{x^6} + 2 = 2704$

$\displaystyle \Rightarrow x^6 + \frac{1}{x^6} = 2702$

$\displaystyle \\$

(xxxii) $\displaystyle \Big( x + \frac{1}{x} \Big)^2 = 3$ find $\displaystyle x^3 + \frac{1}{x^3}$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^2 =3$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big) = \pm \sqrt{3}$

If $\displaystyle x + \frac{1}{x} = \sqrt{3}$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = (\sqrt{3})^3 -3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0$

Similarly, if $\displaystyle x + \frac{1}{x} = -\sqrt{3}$

Then $\displaystyle x^3 + \frac{1}{x^3} = (-\sqrt{3})^3 - 3(-\sqrt{3}) = 0$

$\displaystyle \\$

(xxxiii) If $\displaystyle x + \frac{1}{x} = k$ prove that $\displaystyle x^3 + \frac{1}{x^3} = k(k^2-3)$

If $\displaystyle x + \frac{1}{x} = k$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = (k)^3 -3(k) = k(k^2-3)$ Hence proved.

$\displaystyle \\$

(xxxiv) If $\displaystyle a+b = 10$ and $\displaystyle ab=16$, find the value of $\displaystyle a^2-ab+b^2$ and $\displaystyle a^2+ab+b^2$

Given: $\displaystyle a+b = 10$ and $\displaystyle ab=16$

$\displaystyle (a+b)^2 = a^2 + 2ab + b^2$

$\displaystyle \Rightarrow a^2 + ab + b^2 = (a+b)^2 -ab = 10^2 - 16 = 84$

also $\displaystyle \Rightarrow a^2 - ab + b^2 = (a+b)^2 - 3ab = 10^2 - 3 \times 16 = 100 - 48 = 52$

$\displaystyle \\$

(xxxv) If $\displaystyle a+b = 8$ and $\displaystyle ab=6$, find the value of $\displaystyle a^3+b^3$

Given: $\displaystyle a+b = 8$ and $\displaystyle ab=6$

$\displaystyle (a+b)^3 = a^3 + b^3 + 3ab(a+b)$

$\displaystyle \Rightarrow a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 8^3 - 3 \times 6 \times 8 = 368$

$\displaystyle \\$

(xxxvi) If $\displaystyle a-b = 6$ and $\displaystyle ab=20$, find the value of $\displaystyle a^3-b^3$

Given: $\displaystyle a-b = 6$ and $\displaystyle ab=20$

$\displaystyle (a-b)^3 = a^3 - b^3 - 3ab(a-b)$

$\displaystyle \Rightarrow a^3 - b^3 = (a-b)^3 + 3ab(a-b) = 6^3 + 3 \times 20 \times 6 = 576$

$\displaystyle \\$

Question 5:

(i) If $\displaystyle 3x+2y = 12$ and $\displaystyle xy = 6$, find the value of $\displaystyle 9x^2+4y^2$

$\displaystyle (3x+2y)^2 = 9x^2 + 4y^2 + 12xy$

$\displaystyle \Rightarrow 9x^2 + 4y^2 = 12^2 - 12 \times 6 = 72$

$\displaystyle \\$

$\displaystyle \text{(ii) } 4x^2+y^2 = 40$ and $\displaystyle xy = 6$, find the value of $\displaystyle 2x+y$

$\displaystyle (2x+y)^2 = 4x^2 + y^2 + 4xy = 40 + 4 \times 16 = 64$

$\displaystyle \Rightarrow 2x+y = \pm 8$

$\displaystyle \\$

(iii) If $\displaystyle a^2+b^2+c^2 - ab - bc - ca = 0$, prove that $\displaystyle a = b = c$

Given $\displaystyle a^2+b^2+c^2 - ab - bc - ca = 0$

$\displaystyle \Rightarrow 2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0$

$\displaystyle \Rightarrow (a^2 -2ab+b^2) +(b^2 -2bc + c^2) + (c^2 -2ca + a^2) = 0$

$\displaystyle \Rightarrow (a-b)^2+(b-c)^2 + (c-a)^2 = 0$

$\displaystyle \Rightarrow a-b = 0 \ and \ b -c = 0 \ and \ c-a = 0$

$\displaystyle \Rightarrow a = b \ and \ b = c \ and \ c= a$

$\displaystyle \Rightarrow a = b = c$

$\displaystyle \\$

(iv) If $\displaystyle 9x^2+25y^2 = 181$ and $\displaystyle xy = -6$, find the value of $\displaystyle 3x+5y$

$\displaystyle (3x+5y)^2 = 9x^2 +25y^2 + 30xy = 181-180 = 1$

$\displaystyle \Rightarrow 3x+5y = \pm 1$

$\displaystyle \\$

(v) If $\displaystyle 2x+3y=8$ and $\displaystyle xy = 2$, find the value of $\displaystyle 4x^2+9y^2$

$\displaystyle (2x+3y)^2 = 4x^2 +9y^2+12xy$

$\displaystyle \Rightarrow 4x^2 + 9y^2 = 8^2 - 12 \times 2 = 64 -24 =40$

$\displaystyle \\$

(vi) If $\displaystyle 3x-7y=10$ and $\displaystyle xy = -1$, find the value of $\displaystyle 9x^2+49y^2$

$\displaystyle (3x+7y)^2 = 9x^2 + 49y^2 + 42xy$

$\displaystyle \Rightarrow 9x^2 + 49y^2 = 10^2 -42(-1) = 142$

$\displaystyle \\$

(vii) Prove that $\displaystyle a^2+b^2+c^2 - ab - bc - ca = 0$ is always non negative for all values of $\displaystyle a, b, \ and \ c$.

To prove that $\displaystyle a^2+b^2+c^2 - ab - bc - ca = 0$ is always non-negative or To prove that $\displaystyle 2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0$ is always non-negative

$\displaystyle \Rightarrow (a-b)^2 + (b-c)^2 + (c-a)^2$ which is always not negative.

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(viii) If $\displaystyle a+b = 8$ and $\displaystyle a-b=6$, find $\displaystyle (a^2+b^2)$

$\displaystyle (a+b)^2 = a^2 + b^2 + 2ab$

$\displaystyle \Rightarrow a^2 +b^2 = 8^2 - 2 \times 6 = 64-12 = 52$

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(ix) If $\displaystyle a+b = 6$ and $\displaystyle a-b=4$, find $\displaystyle ab$

$\displaystyle (a+b)^2 = (a-b)^2 +4ab$

$\displaystyle \Rightarrow 4ab = (a+b)^2 - (a-b)^2 = 36 - 24 = 12$

$\displaystyle \Rightarrow ab = 3$

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(x) If $\displaystyle 2x-y+z=0$, prove that $\displaystyle 4x^2-y^2+z^2+4xz=0$

Given: $\displaystyle 2x-y+z=0$

$\displaystyle \Rightarrow 2x+z = y$

$\displaystyle \Rightarrow 4x^2 + z^2 + 4xz = y^2$

$\displaystyle \Rightarrow 4x^2-y^2+z^2+4xz=0$ \$

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(xi) If $\displaystyle 2x+3y=13$ and $\displaystyle xy=6$, find the value of $\displaystyle 8x^3+27y^3$

Given: $\displaystyle 2x+3y=13$ and $\displaystyle xy=6$

$\displaystyle (2x+3y)^3 = 8x^3+27y^3 + 3 \times 2x \times 3y (2x+3y)$

$\displaystyle \Rightarrow 8x^3+27y^3 = 13^3 - 3 \times 6 \times 6 \times 13 = 793$

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(xii) If $\displaystyle 3x-2y=13$ and $\displaystyle xy = 12$, find the value of $\displaystyle 27x^3-8y^3$

Given: $\displaystyle 3x-2y=13$ and $\displaystyle xy = 12$

$\displaystyle (3x-2y)^3 = 27x^3-8y^3 - 3 \times 3x \times 2y (3x-2y)$

$\displaystyle \Rightarrow 27x^3-8y^3 = 13^3 + 3 \times 6 \times 12 \times 13 = 5005$

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(xiii) If $\displaystyle 4x-5z=16$ and $\displaystyle xz=12$, find the value of $\displaystyle 64x^3-125z^3$

Given: $\displaystyle 4x-5z=16$ and $\displaystyle xz=12$

$\displaystyle (4x-5y)^3 = 64x^3-125z^3 - 3 \times 4x \times 5y (4x-5y)$

$\displaystyle \Rightarrow 64x^3-125z^3 = 16^3 + 3 \times 20 \times 12 \times 16 = 15616$

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(xiv) If $\displaystyle \frac{x^2+1}{x} = 4$, find the value of $\displaystyle 2x^3+ \frac{2}{x^3}$

Given: $\displaystyle \frac{x^2+1}{x} = 4$

$\displaystyle \Rightarrow x + \frac{1}{x} = 4$

$\displaystyle x + \frac{1}{x} = 4$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3. x . \frac{1}{x} (x + \frac{1}{x} )$

$\displaystyle \Rightarrow \Big( x + \frac{1}{x} \Big)^3 = x^3 + \frac{1}{x^3} + 3 (x + \frac{1}{x} )$

$\displaystyle x^3 + \frac{1}{x^3} = 4^3 -3(4) = 64-12 = 52$

Therefore $\displaystyle 2 \Big( x^3 + \frac{1}{x^3} \Big) = 52 \times 2 = 108$

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(xv) If $\displaystyle x + y + z = 8$ and $\displaystyle xy+yz+zx = 20$, find the value of $\displaystyle x^3+y^3+z^3-3xyz$

We know $\displaystyle x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2+y^2+z^2 -xy-yz-zx)$
$\displaystyle \Rightarrow x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2zx-3xy-3yz-3zx)$
$\displaystyle \Rightarrow x^3+y^3+z^3-3xyz=(x+y+z) \Big( (x+y+z)^2-3(xy+yz+zx) \Big)$
$\displaystyle \Rightarrow x^3 + y^3 + z^3 -3xyz = 8 (8^2 - 3 \times 20) = 8 \times 4 = 32$