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

Answer:

\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

Answer:

\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.

Answer:

\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}

Answer:

(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}

Answer:

(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}

Answer:

(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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

(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}

Answer:

\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}}

Answer:

\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

Answer:

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

Answer:

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

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

\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}

Answer:

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}

Answer:

\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}

Answer:

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}

Answer:

\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)

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

\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

Answer:

\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

Answer:

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

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(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 .

Answer:

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)

Answer:

\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

Answer:

\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

Answer:

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

Answer:

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

Answer:

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

Answer:

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}

Answer:

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

Answer:

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