SELINA Solution Class 9 Expansions Chapter 4 Exercise 4B

Question 1.1

Find the cube of : 3a- 2b

Sol:

( a - b )³ = a³ - 3ab( a - b ) - b³
( 3a - 2b )³ = (3a)³ - 3 x 3a x 2b( 3a - 2b) - (2b)³
                  = 27a³ - 18ab( 3a - 2b ) - 8b³
                  = 27a³ - 54a²b + 36ab² - 8b³

Question 1.2

Find the cube of : 5a + 3b

Sol:

( a + b )³ = a³ + 3ab( a + b ) + b³
( 5a + 3b)³ = (5a)³ + 3 x 5a x 3b( 5a + 3b) + (3b)³
                  = 125a³ + 45ab( 5a + 3b ) + 27b³
                  = 125a³ + 225a²b + 135ab² + 27b³  

Question 1.3

Find the cube of : ${\displaystyle 2a+\frac{1}{2a}}$     ( a ≠ 0 )

Sol:

( a + b )3 = a3 + 3ab( a + b ) + b3
∴ ${\displaystyle {(2a+\frac{1}{2a})}^3={\left(2a\right)}^3+3\times 2a\times \frac{1}{2a}\times (2a+\frac{1}{2a})+{\left(\frac{1}{2a}\right)}^3}$

∴ ${\displaystyle {(2a+\frac{1}{2a})}^3=8a^3+3(2a+\frac{1}{2a})+\frac{1}{{\left(8a\right)}^3}}$

∴ ${\displaystyle {(2a+\frac{1}{2a})}^3=8a^3+6a+\frac{3}{2a}+\frac{1}{{\left(8a\right)}^3}}$

Question 1.4

Find the cube of : ${\displaystyle (3a-\frac{1}{a}) (a\ne 0)}$

Sol:

( a - b )³ = a³ - 3a²b + 3ab² - b³

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

= ${\displaystyle {\left(3a\right)}^3-3\times {\left(3a\right)}^2\times \frac{1}{a}+3.3a{\left(\frac{1}{a}\right)}^2-{\left(\frac{1}{a}\right)}^3}$

= ${\displaystyle 27a^3-3.9a^2.\frac{1}{a}+9a.\frac{1}{a^2}-\frac{1}{a^3}}$

= ${\displaystyle 27a^3-27a+\frac{9}{a}-\frac{1}{a^3}}$.

Question 2

If  a² + ${\displaystyle \frac{1}{a^2}=47}$ and a ≠ 0   find :
(i) ${\displaystyle a+\frac{1}{a}}$
(ii) ${\displaystyle a^3+\frac{1}{a^3}}$

Sol:

(i) ${\displaystyle a+\frac{1}{a}}$

${\displaystyle a^2+\frac{1}{a^2}=47}$

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

⇒ ${\displaystyle {(a+\frac{1}{a})}^2=47+2}$

⇒ ${\displaystyle {(a+\frac{1}{a})}^2=49}$

⇒ ${\displaystyle a+\frac{1}{a}=\pm \sqrt{49}}$

⇒ ${\displaystyle a+\frac{1}{a}=\pm 7}$

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

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

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

⇒ ${\displaystyle a^3+\frac{1}{a^3}={(\pm 7)}^3-3(\pm 7)}$        [ From (1) ]

⇒ ${\displaystyle a^3+\frac{1}{a^3}=\pm 322}$

Question 3

If  ${\displaystyle a^2+\frac{1}{a^2}}$ = 18; a ≠ 0 find :

(i) ${\displaystyle a-\frac{1}{a}}$

(ii) ${\displaystyle a^3-\frac{1}{a^3}}$

Sol:

${\displaystyle a^2+\frac{1}{a^2}}$ = 18`

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

⇒ ${\displaystyle {(a-\frac{1}{a})}^2=18-2}$ 

⇒ ${\displaystyle {(a-\frac{1}{a})}^2=16}$

⇒ ${\displaystyle a-\frac{1}{a}=\pm \sqrt{16}}$

⇒ ${\displaystyle a-\frac{1}{a}=\pm 4}$               ...(1)

(ii) ${\displaystyle {(a-\frac{1}{a})}^3=a^3-\frac{1}{a^3}-3(a-\frac{1}{a})}$

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

⇒ ${\displaystyle a^3-\frac{1}{a^3}={(\pm 4)}^3+3(\pm 4)}$           [ From(1) ]

⇒ ${\displaystyle a^3-\frac{1}{a^3}=\pm 76}$ 

Question 4

If ${\displaystyle a+\frac{1}{a}}$ = p and a ≠ 0 ; then show that :

${\displaystyle a^3+\frac{1}{a^3}=p(p^2-3)}$ 

Sol:

Given that ${\displaystyle a+\frac{1}{a}}$ = p            .....(1)

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

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

⇒ ${\displaystyle a^3+\frac{1}{a^3}={\left(p\right)}^3-3\left(p\right)}$             [ From(1) ]

⇒ ${\displaystyle a^3+\frac{1}{a^3}=p(p^2-3)}$

Question 5

If a + 2b = 5; then show that : a³ + 8b³ + 30ab = 125.

Sol:

Given that a + 2b = 5 ;
We need to find a³ + 8b³ + 30ab :
Now consider the cube of a + 2b :
( a + 2b )³ = a³ + (2b)³ + 3 x a x 2b x ( a + 2b )
                  = a³ + 8b³ + 6ab x ( a + 2b )
             53 = a³ + 8b³ + 6ab x 5       [ ∵ a + 2b = 5 ]
            125 = a³ + 8b³ + 30ab
Thus the value of a³ + 8b³ + 30ab is 125.

Question 6

If ${\displaystyle {(a+\frac{1}{a})}^2=3\text{and a ≠ 0; then show}:a^3+a^{\frac{1}{3}}=0}$

Sol:

Given that ${\displaystyle {(a+\frac{1}{a})}^2=3}$

⇒ ${\displaystyle a+\frac{1}{a}=\pm \sqrt{3}}$      ....(1)

We need to find ${\displaystyle a^3+\frac{1}{a^3}}$ :
Consider the identity,

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

⇒ ${\displaystyle a^3+\frac{1}{a^3}={(\pm \sqrt{3})}^3-3(\pm \sqrt{3})}$     [ From (1) ]

⇒ ${\displaystyle a^3+\frac{1}{a^3}=\pm 3\sqrt{3}-3(\pm \sqrt{3})}$

⇒ ${\displaystyle a^3+\frac{1}{a^3}=0}$

Question 7

If a + 2b + c = 0; then show that : a³ + 8b³ + c³ = 6abc.

Sol:

Given that a + 2b + c = 0;
⇒ a + 2b = -c          ....(1)
Now consider the expansion of ( a + 2b )³ :
                                                    ( a + 2b )³ = ( - c )³
a³ + (2b)³ + 3 x a x 2b x ( a + 2b ) = -c³

⇒         a³ + 8b³ + 3 x a x 2b x (-c) = -c³          [ from (1) ]

⇒                           a³ + 8b³ - 6abc = -c³   

⇒                              a³ + 8b³ - c³ = 6abc

Hence proved.

Question 8.1

Use property to evaluate : 13³ + (-8)³ + (-5)³

Sol:

Property is if a + b + c = 0 then a³ + b³ + c³ = 3abc
a = 13, b = -8 and c = -5
13³ + (-8)³ + (-5)³ = 3(13)(-8)(-5) = 1560

Question 8.2

Use property to evaluate : 7³ + 3³ + (-10)³

Sol:

Property is if a + b + c = 0 then a³ + b³ + c³ = 3abc
a = 7, b = 3, c = -10
7³ + 3³ + (-10)³ = 3(7)(3)(-10) = -630

Question 8.3

Use property to evaluate : 9³ - 5³ - 4³

Sol:

Property is if a + b + c = 0 then a³ + b³ + c³ = 3abc
a = 9, b = -5, c = -4
9³ - 5³ - 4³ = 9³ + (-5)³ + (-4)³ = 3(9)(-5)(-4) = 540

Question 8.4

Use property to evaluate : 38³ + (-26)³ + (-12)³

Sol:

Property is if a + b + c = 0 then a³ + b³ + c³ = 3abc
a = 38, b = -26, c = -12
38³ + (-26)³ + (-12)³ = 3(38)(-26)(-12) = 35568

Question 9.1

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 3 ; find : 
${\displaystyle a^2+\frac{1}{a^2}}$

Sol:

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

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

${\displaystyle a^2+\frac{1}{a^2}}$  = 9 + 2 = 11

Question 9.2

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 3 ; Find :
${\displaystyle a^3-\frac{1}{a^3}}$

Sol:

${\displaystyle a-\frac{1}{a}}$ = 3...............(Given)

Taking cube on both sides,

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

${\displaystyle a^3+\frac{1}{a^3}-3(a-\frac{1}{a})=27}$..............[(a - b)³ = a³ - b³ -3ab(a - b)]

${\displaystyle a^3+\frac{1}{a^3}-3\times 3=27..............[a-\frac{1}{a}=3]}$

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

${\displaystyle a^3+\frac{1}{a^3}}$ = 27 + 9

${\displaystyle a^3+\frac{1}{a^3}}$ = 36.

Question 10.1

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 4 ; find : ${\displaystyle (a^2+\frac{1}{a^2})}$

Sol:

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

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

⇒ ${\displaystyle a^2+\frac{1}{a^2}={\left(4\right)}^2+2 [\because a-\frac{1}{a}=4]}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}}$ = 18

Question 10.2

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 4 ; find : ${\displaystyle (a^4+\frac{1}{a^4})}$

Sol:

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

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

⇒ ${\displaystyle a^2+\frac{1}{a^2}={\left(4\right)}^2+2 [\because a-\frac{1}{a}=4]}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}}$ = 18                        ...(1)

We know that,

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

                       = `(18)^2 - 2             [ From(1) ]
                       = 324 - 2

⇒ ${\displaystyle a^4+\frac{1}{a^4}=322}$

Question 10.3

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 4 ; find : ${\displaystyle (a^3-\frac{1}{a^3})}$

Sol:

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

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

⇒ ${\displaystyle (a^3-\frac{1}{a^3})={\left(4\right)}^3+3\left(4\right)}$         [ ∵ ${\displaystyle a-\frac{1}{a}=4}$ ]

⇒ ${\displaystyle (a^3-\frac{1}{a^3})=64+12}$ 

⇒ ${\displaystyle (a^3-\frac{1}{a^3})=76}$ 

Question 11

If X ≠ 0 and X + ${\displaystyle \frac{1}{\text{X}}}$ = 2 ; then show that :

${\displaystyle x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}}$

Sol:

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

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

⇒ ${\displaystyle x^2+\frac{1}{x^2}={\left(2\right)}^2-2 [\because x+\frac{1}{x}=2]}$

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

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

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

⇒ ${\displaystyle x^3+\frac{1}{x^3}={\left(2\right)}^3-3\left(2\right) [\because x+\frac{1}{x}=2]}$

⇒ ${\displaystyle x^3+\frac{1}{x^3}=8-6}$   

⇒ ${\displaystyle x^3+\frac{1}{x^3}=2}$                ...(2)      

We know that

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

                       = ${\displaystyle {\left(2\right)}^2-2}$               [ from (1) ]
                       = 4 - 2
⇒ ${\displaystyle x^4+\frac{1}{x^4}=2}$                ...(3)

Thus from equations (1), (2) and (3), we have

${\displaystyle x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}}$

Question 12

If 2x - 3y = 10 and xy = 16; find the value of 8x³ - 27y³.

---

Sol:

Given that 2x - 3y = 10, xy = 16

∴ (2x - 3y)³ = (10)³

⇒ 8x³ - 27y³ - 3 (2x) (3y) (2x - 3y) = 1000 

⇒ 8x³ - 27 y³ -18xy (2x - 3y) = 1000

⇒ 8x³ - 27 y³ - 18 (16) (10)  = 1000

⇒ 8x³ - 27 y³ - 2880 = 1000

⇒8x³ - 27 y³ = 1000 + 2880

⇒ 8x³ - 27 y³ =3880

Question 13.1

Expand : (3x + 5y + 2z) (3x - 5y + 2z)

Sol:

(3x + 5y + 2z) (3x - 5y + 2z)

= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}

= (3x + 2z)² - (5y)²

{since (a + b) (a - b) = a² - b²}

= 9x² + 4z² + 2 × 3x × 2z - 25y²

= 9x² + 4z² + 12xz - 25y²

= 9x² + 4z² - 25y² + 12xz

Question 13.2

Expand : (3x - 5y - 2z) (3x - 5y + 2z)

Sol:

(3x - 5y - 2z) (3x - 5y + 2z)

= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}

= (3x - 5y)² - (2z)²{since(a + b) (a - b) = a² - b²}

= 9x² + 25y² - 2 × 3x × 5y - 4z²

= 9x² + 25y²- 30xy - 4z²

= 9x² +25y² - 4z² - 30xy

Question 14

The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes

Sol:

Given sum of two numbers is 9 and their product is 20.
Let the numbers be a and b.
a + b = 9
ab = 20

Squaring on both sides gives
(a+b)² = 9²
a² + b² + 2ab = 81
a² + b² + 40 = 81
So sum of squares is 81 - 40 = 41

Cubing on both sides gives
(a + b)³ = 9³
a³ + b³ + 3ab(a + b) = 729
a³ + b³ + 60(9) = 729
a³ + b³ = 729 - 540 = 189
So the sum of cubes is 189.

Question 15

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.

Sol:

Given x - y = 5 and xy = 24 (x>y)
(x + y)² = (x - y)² + 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.

Cubing on both sides gives
(x - y)³ = 5³
x³ - y³ - 3xy(x - y) = 125
x³ - y³ - 72(5) = 125
x³ - y³= 125 + 360 = 485
So, difference of their cubes is 485.

Cubing both sides, we get
(x + y)³ = 11³
x³ + y³ + 3xy(x + y) = 1331
x³ + y³ = 1331 - 72(11) = 1331 - 792 = 539
So, sum of their cubes is 539.

Question 16

If 4x² + y² = a and xy = b, find the value of 2x + y.

Sol:

xy = ab                                                   ….(i)

4x² + y² = a                                            ….(ii)

Now, (2x + y)² = (2x)² + 4xy + y²

= 4x² + y² + 4xy

= a + 4b                                                 ….[From (i) and (ii)]

⇒ 2x + y = ${\displaystyle \pm \sqrt{a+4b}}$