Question 1: Expand / Simplify the following:

(i) $(3x+4y)^2$

Answer:

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

$\\$

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

Answer:

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

$\\$

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

Answer:

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

$\\$

(iv) $(x-1)(x+1)(x^2+1)(x^4+1)$

Answer:

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

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

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

$= (x^8-1)$

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

Let $2x+5 = a$

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

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

$= a^2 + 7a + 12$

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

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

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

$\\$

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

Answer:

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

$\\$

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

Answer:

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

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

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

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

Answer:

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

$\\$

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

Answer:

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

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

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

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

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

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

$\\$

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

Answer:

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

$\Rightarrow a + b = 2x$

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

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

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

$\\$

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

Answer:

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

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

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

Therefore $(2x-5y)^3-(2x+5y)^3$

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

$\\$

Question 2: Evaluate the following identities:

(i) $103 \times 97$

Answer:

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

$\\$

(ii) $(97)^2$

Answer:

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

$\\$

(iii) $0.54 \times 0.54 -0.46 \times 0.46$

Answer:

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

$\\$

(iv) $(0.98)^2$

Answer:

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

$\\$

(v) $991 \times 1009$

Answer:

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

$\\$

(vi) $(1002)^3$

Answer:

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

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

$= 1000000000+8 + 6012000= 1006012008$

$\\$

(vii) $(999)^3$

Answer:

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

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

$= 1000000000-1-299700 = 997002999$

$\\$

(viii) $(103)^3$

Answer:

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

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

$= 1000000+27+92700 = 1092727$

$\\$

(ix) $(10.4)^3$

Answer:

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

$= 1000 + 0.064+124.8 = 1124.864$

$\\$

(x) $(99)^3$

Answer:

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

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

$= 1000000-1-29700=970299$

$\\$

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

Answer:

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

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

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

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

$= 662662$

$\\$

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

Answer:

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

$104^3+96^3$

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

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

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

$= 2009600$

$\\$

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

Answer:

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

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

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

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

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

$\\$

Question 3:

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

Answer:

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

$a^2 + b^2 = 85$

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

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

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

$\Rightarrow ab = 18$

$\\$

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

Answer:

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

$x^2 + y^2 = 29$

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

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

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

$\Rightarrow xy = 10$

$\\$

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

Answer:

$a+ b = 7$

$a^3 + b^3 = 133$

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

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

$21ab = 133 - 49 = 84$

$\Rightarrow ab = 4$

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

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

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

$\\$

Question 4:

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

Answer:

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

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

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

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

(b)  $x^2 +$ $\frac{1}{x^2}$ $= 34$

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

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

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

$\\$

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

Answer:

(a)  $x =$ $\frac{1}{x - 2\sqrt{3}}$

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

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

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

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

(b)  Since $(a+b)^2 - (a-b)^2 = 4ab$

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

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

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

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

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

(c)  $x-$ $\frac{1}{x}$ $= 2\sqrt{3}$

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

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

$\\$

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

Answer:

(a)  $x^2+$ $\frac{1}{x^2}$ $= 27$

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

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

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

(b)  $x^2+$ $\frac{1}{x^2}$ $= 27$

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

Now

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

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

$\\$

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

Answer:

(a)  $a =$ $\frac{1}{a-5}$

$a^2 - 5a = 1$

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

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

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

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

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

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

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

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

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

$\\$

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

Answer:

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

$a^2+1 = 3a$

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

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

(a)  $a +$ $\frac{1}{a}$ $= 3$

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

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

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

(b)  $a +$ $\frac{1}{a}$ $= 3$

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

Given $x^2 +$ $\frac{1}{x^2}$ $= 7$

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

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

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

If  $x +$ $\frac{1}{x}$ $= 3$

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

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

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

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

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

$\\$

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

Answer:

Given $x^2 +$ $\frac{1}{x^2}$ $= 83$

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

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

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

If  $x -$ $\frac{1}{x}$ $= 9$

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

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

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

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

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

$\\$

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

Answer:

Given $x^4 +$ $\frac{1}{x^4}$ $= 47$

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

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

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

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

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

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

If  $x +$ $\frac{1}{x}$ $= 3$

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

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

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

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

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

$\\$

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

Answer:

Given $a+b = 10$ and $ab = 21$

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

If  $x +$ $\frac{1}{x}$ $= 5$

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

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

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

$\\$

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

Answer:

If  $x -$ $\frac{1}{x}$ $= 7$

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

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

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

$\\$

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

Answer:

Given $x^2 +$ $\frac{1}{x^2}$ $= 51$

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

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

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

If  $x -$ $\frac{1}{x}$ $= 7$

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

Now $x +$ $\frac{1}{x}$ $= \pm 10$

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

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

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

If  $x -$ $\frac{1}{x}$ $= 3$

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

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

If  $x +$ $\frac{1}{x}$ $= 4$

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

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

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

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

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

$\\$

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

Answer:

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

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

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

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

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

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

$\\$

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

Answer:

Given $x =$ $\frac{1}{4-x}$

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

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

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

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

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

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

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

$\\$

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

Answer:

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

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

If  $x +$ $\frac{1}{x}$ $= \sqrt{3}$

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

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

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

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

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

$\\$

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

Answer:

If  $x +$ $\frac{1}{x}$ $= k$

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

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

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

$\\$

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

Answer:

Given: $a+b = 10$ and $ab=16$

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

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

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

$\\$

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

Answer:

Given: $a+b = 8$ and $ab=6$

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

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

$\\$

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

Answer:

Given: $a-b = 6$ and $ab=20$

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

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

$\\$

Question 5:

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

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

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

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

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

$\Rightarrow a = b = c$

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

$\Rightarrow ab = 3$

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

Answer:

Given: $2x-y+z=0$

$\Rightarrow 2x+z = y$

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

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

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

Answer:

Given: $2x+3y=13$ and $xy=6$

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

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

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

Answer:

Given: $3x-2y=13$ and $xy = 12$

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

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

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

Answer:

Given: $4x-5z=16$ and $xz=12$

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

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

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

Answer:

Given: $\frac{x^2+1}{x}$ $= 4$

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

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

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

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

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

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

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

Answer:

We know $x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2+y^2+z^2 -xy-yz-zx)$

$\Rightarrow x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2zx-3xy-3yz-3zx)$

$\Rightarrow x^3+y^3+z^3-3xyz=(x+y+z) \Big( (x+y+z)^2-3(xy+yz+zx) \Big)$

$\Rightarrow x^3 + y^3 + z^3 -3xyz = 8 (8^2 - 3 \times 20) = 8 \times 4 = 32$