SELINA Solution Class 9 Expansions Chapter 4 Exercise 4E

Question 1.1

Simplify : ( x + 6 )( x + 4 )( x - 2 )

Sol:

Using identity :
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

(x + 6)(x + 4)(x - 2)
= x³ + (6 + 4 - 2)x² + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x³ + 8x² + (24 - 8 - 12)x - 48
= x³ + 8x² + 4x - 48

Question 1.2

Simplify : ( x - 6 )( x - 4 )( x + 2 )

Sol:

Using identity : (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

(x - 6)(x - 4)(x + 2)
= x³ + (-6 - 4 + 2)x² + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x³ - 8x² + (24 - 8 - 12)x + 48
= x³ - 8x² + 4x + 48

Question 1.3

Simplify : ( x - 6 )( x - 4 )( x - 2 )

Sol:

Using identity :
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

( x - 6 )( x - 4 )( x - 2 )
= x³ + (-6 - 4 - 2)x² + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x³ - 12x² + (24 + 8 + 12)x - 48
= x³ - 12x² + 44x - 48

Question 1.4

Simplify : ( x + 6 )( x - 4 )( x - 2 )

Sol:

Using identity :
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

 

(x + 6)(x - 4)(x - 2)
= x³ + (6 - 4 - 2)x² + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x³ - 0x² + (-24 + 8 - 12)x + 48
= x³ - 28x + 48 

Question 2.1

Simplify using following identity : ${\displaystyle (a\pm b)(a^2\pm ab+b^2)=a^3\pm b^3}$
( 2x + 3y )( 4x² + 6xy + 9y² )

Sol:

( 2x + 3y )( 4x² + 6xy + 9y² )
= ( 2x + 3y )[ (2x)² - (2x)(3y) + (3y)² ]
= (2x)³ + (3y)³
= 8x³ + 27y³

Question 2.2

Simplify using following identity : ${\displaystyle (a\pm b)(a^2\pm ab+b^2)=a^3\pm b^3}$
${\displaystyle (3x-\frac{5}{x})(9x^2+15+\frac{25}{x^2})}$

Sol:

${\displaystyle (3x-\frac{5}{x})(9x^2+15+\frac{25}{x^2})}$

=${\displaystyle (3x-\frac{5}{x})[{\left(3x\right)}^2+\left(3x\right)\left(\frac{5}{x}\right)+{\left(\frac{5}{x}\right)}^2]}$

= ${\displaystyle {\left(3x\right)}^3-{\left(\frac{5}{x}\right)}^3}$

= ${\displaystyle 27x^3-\frac{125}{x^3}}$

Question 2.3

Simplify using following identity : ${\displaystyle (a\pm b)(a^2\pm ab+b^2)=a^3\pm b^3}$
${\displaystyle (\frac{a}{3}-3b)(\frac{a^2}{9}+ab+9b^2)}$

Sol:

${\displaystyle (\frac{a}{3}-3b)(\frac{a^2}{9}+ab+9b^2)}$

= ${\displaystyle (\frac{a}{3}-3b)[{\left(\frac{a}{3}\right)}^2+\left(\frac{a}{3}\right)\left(3b\right)+{\left(3b\right)}^2]}$

= ${\displaystyle {\left(\frac{a}{3}\right)}^3-{\left(3b\right)}^3}$

= ${\displaystyle \frac{a^3}{27}-27b^3}$

Question 3.1

Using suitable identity, evaluate (104)³

Sol:

Using identity: (a ± b)³ = a³ ± b³ ± 3ab(a ± b)
(104)³ = (100 + 4)³
= (100)³ + (4)³ + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864

Question 3.2

Using suitable identity, evaluate (97)³ 

Sol:

(97)³ = (100 - 3)³
= (100)³ - (3)³ - 3 × 100 × 3(100 - 3)
= 1000000 - 27 - 900 × 97
= 1000000 - 27 - 87300
= 912673

Question 4

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

Sol:

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

If a + b + c = 0, then a³ + b³ + c³ = 3abc
Now, x² - y² + y² - z² + z² - x² = 0

⇒  ( x² - y² )³ + ( y² - z² )³ + ( z² - x² )³ = 3( x² - y² )( y² - z² )( z² -x² )                                         ......(1)

And, x - y + y - z + z - x = 0

⇒  ( x - y )³ + ( y - z )³ + ( z - x )³ = 3( x - y )( y - z )( z -x )                                                          ........(2)
Now,
${\displaystyle \frac{{(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}}$

= ${\displaystyle \frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}}$                            .....[From (1) and (2)] 

= ( x + y )( y + z )( z + x ) 

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

If a + b + c = 0, then a³ + b³ + c³ = 3abc
Now, x² - y² + y² - z² + z² - x² = 0

⇒  ( x² - y² )³ + ( y² - z² )³ + ( z² - x² )³ = 3( x² - y² )( y² - z² )( z² -x² )                                         ......(1)

And, x - y + y - z + z - x = 0

⇒  ( x - y )³ + ( y - z )³ + ( z - x )³ = 3( x - y )( y - z )( z -x )                                                          ........(2)
Now,
${\displaystyle \frac{{(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}}$

= ${\displaystyle \frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}}$                            .....[From (1) and (2)] 

= ( x + y )( y + z )( z + x ) 

Question 5.1

Evaluate :
${\displaystyle \frac{0.8\times 0.8\times 0.8+0.5\times 0.5\times 0.5}{0.8\times 0.8-0.8\times 0.5+0.5\times .5}}$

Sol:

${\displaystyle \frac{0.8\times 0.8\times 0.8+0.5\times 0.5\times 0.5}{0.8\times 0.8-0.8\times 0.5+0.5\times .5}}$

Let 0.8 = a and 0.5 = b
Then, the given expression becomes
${\displaystyle \frac{a\times a\times a+b\times b\times b}{a\times a-a\times b+b\times b}}$

= ${\displaystyle \frac{a^3+b^3}{a^2-ab+b^2}}$

= ${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}}$

= a + b
= 0.8 + 0.5
= 1.3

Question 5.2

Evaluate :
${\displaystyle \frac{1.2\times 1.2+1.2\times 0.3+0.3\times 0.3}{1.2\times 1.2\times 1.2- 0.3\times 0.3\times 0.3}}$

Sol:

${\displaystyle \frac{1.2\times 1.2+1.2\times 0.3+0.3\times 0.3}{1.2\times 1.2\times 1.2- 0.3\times 0.3\times 0.3}}$

Let 1.2 = a and 0.3 = b
Then, the given expression becomes
${\displaystyle \frac{a\times a+a+b+b\times b}{a\times a\times a-b\times b\times b}}$

= ${\displaystyle \frac{a^2+ab+b^2}{a^3-b^3}}$

= ${\displaystyle \frac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)}}$

= ${\displaystyle \frac{1}{a-b}}$

= ${\displaystyle \frac{1}{1.2-0.3}}$

= ${\displaystyle \frac{1}{0.9}}$

= ${\displaystyle \frac{10}{9}}$

= ${\displaystyle 1\frac{1}{9}}$

Question 6

If a - 2b + 3c = 0; state the value of a³ - 8b³ + 27c³.

Sol:

a³ - 8b³ + 27c³ = a³ + (-2b)³ + (3c)³
Since a - 2b + 3c = 0, we have
a³ - 8b³ + 27c³ = a³ + (-2b)³ + (3c)³
= 3(a)( -2b)(3c)
= -18abc

Question 7

If x + 5y = 10; find the value of x³ + 125y³ + 150xy - 1000.

Sol:

x + 5y = 10
⇒ (x + 5y)³ = 10³
⇒ x³ + (5y)³ + 3(x)(5y)(x + 5y) = 1000
⇒ x³ + (5y)³ + 3(x)(5y)(10) = 1000
= x³ + (5y)³ + 150xy = 1000
= x³ + (5y)³ + 150xy - 1000 = 0

Question 8

If x = 3 + 2√2, find :
(i) ${\displaystyle \frac{1}{x}}$

(ii) ${\displaystyle x-\frac{1}{x}}$

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

(iv) ${\displaystyle x^3-\frac{1}{x^3}}$

Sol:

x = 3 + 2√2
(i) ${\displaystyle \frac{1}{x}=\frac{1}{3+2\surd 2}}$

= ${\displaystyle \frac{1}{3+2\surd 2}\times \frac{3-2\surd 2}{3-2\surd 2}}$

= ${\displaystyle \frac{3-2\surd 2}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}}$

= ${\displaystyle \frac{3-2\surd 2}{9-8}}$

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

(ii) ${\displaystyle x-\frac{1}{x}=(3+2\sqrt{2})-(3-2\sqrt{2})}$   ...[From(2)] 
                 = ${\displaystyle 3+2\sqrt{2}-3+2\sqrt{2}}$
∴ ${\displaystyle x-\frac{1}{x}=4\sqrt{2}}$              ....(2)

(iii) ${\displaystyle {(x-\frac{1}{x})}^3={\left(4\sqrt{2}\right)}^3}$
                          = 64 x 2√2
                          = 128√2

(iv) ${\displaystyle x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})}$
                           = 128√2 + 3(4√2)
                           = 128√2 + 12√2
                           = 140√2

Question 9

If a + b = 11 and a² + b² = 65; find a³ + b³.

Sol:

a + b = 11 and a2 + b2 = 65
Now, (a+b)² = a² + b² + 2ab
⇒ (11)² = 65 + 2ab
⇒ 121 = 65 + 2ab
⇒  2ab = 56
⇒  ab = 28

a³ + b³ = ( a + b )( a² - ab + b2)
             = (11)(65 - 28)
             = 11 x 37
             = 407

Question 10

Prove that :  x²+ y² + z² - xy - yz - zx  is always positive.

Sol:

x² + y² + z² - xy - yz - zx

= 2(x² + y² + z² - xy - yz - zx)

= 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

= x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2zx

= (x² + y² - 2xy) + (z² + x² - 2zx) + (y² + z² - 2yz)

= (x - y)² + (z - x)² + (y - z)²

Since square of any number is positive, the given equation is always positive.

Question 11.1

Find : (a + b)(a + b)

Sol:

(a + b)(a + b) = (a + b)²
= a × a + a × b + b × a + b × b
= a² + ab + ab + b²
= a² + b² + 2ab

Question 11.2

Find : (a + b)(a + b)(a + b)

Sol:

(a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a² + ab + ab + b²)(a + b)
= (a² + b² + 2ab)(a + b)
= a² × a + a² × b + b² × a + b² × b + 2ab × a + 2ab × b
= a³ + a² b + ab² + b³ + 2a²b + 2ab²
= a³ + b³ + 3a²b + 3ab²

Question 11.3

Find : (a - b)(a - b)(a - b)

Sol:

(a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a² + ab + ab + b²)(a + b)
= (a² + b² + 2ab)(a + b)
= a² × a + a² × b + b² × a + b² × b + 2ab × a + 2ab × b
= a³ + a² b + ab² + b³ + 2a²b + 2ab²
= a³ + b³ + 3a²b + 3ab²

replacing b by -b, we get
= a³ + (-b)³ + 3a²(-b) + 3a(-b)²
= a³ - b³ - 3a²b + 3ab²