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

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(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  

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(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}

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(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)

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(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)

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(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}

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(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 

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(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

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(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

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(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

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(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}

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(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

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(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

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(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

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(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

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(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)

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

Answer:

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

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(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

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(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

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(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

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(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

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(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}

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(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

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(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

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(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

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(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 

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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

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(ii) (97)^2

Answer:

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

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(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

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(iv) (0.98)^2

Answer:

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

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(v) 991 \times 1009

Answer:

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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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}

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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

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(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

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(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

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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

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(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}

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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}

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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}

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(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

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(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

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(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.

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(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

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(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 

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(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 

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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

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(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

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(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

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(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 

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(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 

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(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 

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(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