# RD Sharma Solutions Class 9 Algebraic Identities Exercise 4.3

## RD Sharma Solutions Class 9 Chapter 4 Ex 4.3

Q1. Find the cube of each of the following binomial expression

(a) $(\frac{1}{x} + \frac{y}{3})$

(b) $(\frac{3}{x} – \frac{2}{x^{2}})$

(c) $(2x + \frac{3}{x})$

(d) $(4 – \frac{1}{3x})$

Solution :

(a) As per the question, we have to find the value of; $(\frac{1}{x} + \frac{y}{3})$3

The given expression is in the form of (a + b)3 = a3 + b3 + 3ab(a + b)

Let, a = 1/x , b = y/3

Thus, by (a + b)formula;

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

= $\frac{1}{x ^{3}}$ + $\frac{y ^{3}}{27}$ + 3 * $\frac{1}{x}$ * $\frac{y}{3}$( $\frac{1}{x}$ + $\frac{y}{3}$)

= $\frac{1}{x ^{3}}$ + $\frac{y ^{3}}{27}$ + $\frac{y}{x}$( $\frac{1}{x}$ + $\frac{y}{3}$)

= $\frac{1}{x ^{3}}$ + $\frac{y ^{3}}{27}$ + ($\frac{y}{x}$ * $\frac{1}{x}$) +

($\frac{y}{x}$ * $\frac{y}{3}$)

= $\frac{1}{x ^{3}}$ + $\frac{y ^{3}}{27}$ + $\frac{y}{x ^{2}}$ + $\frac{y ^{2}}{3x}$)

Therefore,

$(\frac{1}{x} + \frac{y}{3})$)3 = $\frac{1}{x ^{3}}$ + $\frac{y ^{3}}{27}$ + $\frac{y}{x ^{2}}$ + $\frac{y ^{2}}{3x}$)

(b) As per the question, we have to find the value of; $(\frac{3}{x} – \frac{2}{x^{2}})$3

The given expression is in the form of (a – b)3 = a3 – b3 – 3ab(a – b)

Let, a = 3/x , b = $\frac{2}{x^{2}}$

Thus, by (a – b)formula

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

= $\frac{27}{x ^{3}}$$\frac{8}{x ^{6}}$ – 3 * $\frac{3}{x}$ * $\frac{2}{x ^{2}}$( $\frac{3}{x}$$\frac{2}{x ^{2}}$)

= $\frac{27}{x ^{3}}$$\frac{8}{x ^{6}}$$\frac{18}{x ^{3}}$($\frac{3}{x}$$\frac{2}{x ^{2}}$)

= $\frac{27}{x ^{3}}$$\frac{8}{x ^{6}}$ – ($\frac{18}{x ^{3}}$ * $\frac{3}{x}$) + ($\frac{18}{x ^{3}}$ * $\frac{2}{x ^{2}}$)

= $\frac{27}{x ^{3}}$$\frac{8}{x ^{6}}$$\frac{54}{x ^{4}}$ + $\frac{36}{x ^{5}}$

Therefore, ($(\frac{3}{x} – \frac{2}{x^{2}})$)3 = $\frac{27}{x ^{3}}$$\frac{8}{x ^{6}}$$\frac{54}{x ^{4}}$ + $\frac{36}{x ^{5}}$

(c)  As per the question, we have to find the value of; $(2x + \frac{3}{x})$3

The given expression is in the form of (a + b)3 = a3 + b3 + 3ab(a + b)

Let, a = 2x , b = 3/x

Thus, by (a + b)formula

= 8x3 + $\frac{27}{x ^{3}}$ + $\frac{18x}{x}$( 2x + $\frac{3}{x}$)

= 8x3 + $\frac{27}{x ^{3}}$ + $\frac{18x}{x}$( 2x + $\frac{3}{x}$)

= 8x3 + $\frac{27}{x ^{3}}$ +

(18 * 2x) + ( 18 * $\frac{3}{x}$)

= 8x3 + $\frac{27}{x ^{3}}$ + 36x $\frac{54}{x}$)

Therefore,

The cube of  $(2x + \frac{3}{x})$3 =  8x3 + $\frac{27}{x ^{3}}$ + 36x $\frac{54}{x}$)

(d) As per the question, we have to find the value of; $(4 – \frac{1}{3x})$3

The given expression is in the form of (a – b)3 = a3 – b3 – 3ab(a – b)

Let, a = 4 , b = 1/3x

Thus, by (a – b)formula;

$(4 – \frac{1}{3x})$3 = 43 – ($\frac{1}{3x}$)3 – 3(4)($\frac{1}{3x}$)(4 – $\frac{1}{3x}$)

= 64 – $\frac{1}{27x ^{3}}$$\frac{12}{3x}$( 4 – $\frac{1}{3x}$)

= 64 – $\frac{1}{27x ^{3}}$$\frac{4}{x}$( 4 – $\frac{1}{3x}$)

= 64 – $\frac{1}{27x ^{3}}$ – ($\frac{4}{3x}$ * 4) + ($\frac{4}{3x}$ * $\frac{1}{3x}$)

= 64 – $\frac{1}{27x ^{3}}$$\frac{16}{x}$  + ($\frac{4}{3x ^{2}}$

Therefore,

The cube of $(4 – \frac{1}{3x})$3 = 64 – $\frac{1}{27x ^{3}}$$\frac{16}{x}$  + ($\frac{4}{3x ^{2}})$

Q2. Simplify each of the following

(a) (x + 3)3 + ( x – 3)3

(b) $(\frac{x}{2} + \frac{y}{3})^{3}$$(\frac{x}{2} – \frac{y}{3})^{3}$

(c)  $(x + \frac{2}{x})^{3}$ + $(x – \frac{2}{x})^{3}$

(d) (2x – 5y)3 – (2x + 5y)3

Solution :

(a) (x + 3)3 + ( x – 3)3

The given expression is in the form of a3 + b3 = (a + b)(a2 + b2 – ab)

Say, a = (x + 3) , b = (x – 3)

Now, by (a3 + b3)  formula;

= (x + 3 + x – 3)[ (x + 3)3 + (x – 3)3 – (x + 3)(x – 3)]

= 2x[(x2 + 32 + 2*x*3) + (x2 + 32 – 2*x*3) – (x2 – 32)]

=  2x[(x2 + 9 + 6x) + (x2 + 9 – 6x) – x2 + 9]

= 2x[(x2 + 9 + 6x + x2 – 9 – 6x – x2 + 9)]

= 2x ( x2 + 27)

= 2x3 + 54x

Therefore, the result of (x + 3)3 + ( x – 3)3 is 2x3 + 54x

(b) $(\frac{x}{2} + \frac{y}{3})^{3}$$(\frac{x}{2} – \frac{y}{3})^{3}$

The given expression is in the form of a3 – b3 = (a – b)(a2 + b2 + ab)

Let, a = $(\frac{x}{2} + \frac{y}{3})^{3}$ , b = $(\frac{x}{2} – \frac{y}{3})^{3}$

Now, by (a3 – b3)  formula;

= [($(\frac{x}{2} + \frac{y}{3})^{3}$ – ($(\frac{x}{2} – \frac{y}{3})^{3}$ )][($(\frac{x}{2} + \frac{y}{3})^{3}$)2 ($(\frac{x}{2} – \frac{y}{3})^{3}$)2 – ($(\frac{x}{2} + \frac{y}{3})^{3}$ )($(\frac{x}{2} – \frac{y}{3})^{3}$)

= $(\frac{x}{3} + \frac{y}{3} – \frac{x}{2} + \frac{y}{3})[((\frac{x}{2})^{2} + (\frac{y}{3})^{2} + (\frac{2xy}{6})) + ((\frac{x}{2})^{2} + (\frac{y}{3})^{2} – (\frac{2xy}{6})) + ((\frac{x}{2})^{2} – (\frac{y}{3})^{2}) ]$

= $\frac{2y}{3}[(\frac{x^{2}}{4} + \frac{y^{2}}{9} + \frac{2xy}{6}) + (\frac{x^{2}}{4} + \frac{y^{2}}{9} – \frac{2xy}{6}) + \frac{x^{2}}{4} – \frac{y^{2}}{9}]$

= $\frac{2y}{3}[\frac{x^{2}}{4} + \frac{y^{2}}{9} + \frac{2xy}{6} + \frac{x^{2}}{4} + \frac{y^{2}}{9} – \frac{2xy}{6} + \frac{x^{2}}{4} – \frac{y^{2}}{9}]$

= $\frac{2y}{3}[\frac{x^{2}}{4} + \frac{y^{2}}{9} + \frac{x^{2}}{4} + \frac{x^{2}}{4}]$

= $\frac{2y}{3}[\frac{3x^{2}}{4} + \frac{y^{2}}{9}]$

= $\frac{x^{2}y}{2} + \frac{2y^{3}}{27}$

Therefore, the result of $(\frac{x}{2} + \frac{y}{3})^{3}$$(\frac{x}{2} – \frac{y}{3})^{3}$ = $\frac{x^{2}y}{2} + \frac{2y^{3}}{27}$

(c)  $(x + \frac{2}{x})^{3}$ + $(x – \frac{2}{x})^{3}$

The given expression is in the form of a3 + b3 = (a + b)(a2 + b2 – ab)

Let, a = $(x + \frac{2}{x})^{3}$ , b = $(x – \frac{2}{x})^{3}$

Now, by (a3 + b3)  formula;

= $(x + \frac{2}{x} + x – \frac{2}{x})[(x + \frac{2}{x})^{2} + (x – \frac{2}{x})^{2} – ((x + \frac{2}{x})(x – \frac{2}{x}))]$

= $(2x)[(x^{2} + \frac{4}{x^{2}} + \frac{4x}{x}) + (x^{2} + \frac{4}{x^{2}} – \frac{4x}{x}) – (x^{2}- \frac{4}{x^{2}})$

= $(2x)[(x^{2} + \frac{4}{x^{2}} + \frac{4x}{x} + x^{2} + \frac{4}{x^{2}} – \frac{4x}{x} – x^{2} + \frac{4}{x^{2}})$

= $(2x)[(x^{2} + \frac{4}{x^{2}} +\frac{4}{x^{2}} + \frac{4}{x^{2}})$

= $(2x)[(x^{2} + \frac{12}{x^{2}})$

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

Therefore, the result of $(x + \frac{2}{x})^{3}$ + $(x – \frac{2}{x})^{3}$ = $(2x)[(x^{2} + \frac{12}{x^{2}})$

(d) Given here, (2x – 5y)3 – (2x + 5y)3

The given expression is in the form of a3 – b3 = (a – b)(a2 + b2 + ab)

Let, a = (2x – 5y) , b = (2x + 5y)

Now, by (a3 – b3)  formula;

= (2x – 5y – 2x – 5y)[(2x – 5y)2 + (2x + 5y)2 + ((2x – 5y) * (2x + 5y))]

= (-10y)[(4x2 + 25y2 – 20xy) + (4x2 + 25y2 + 20xy) + 4x2 – 25y2]

= (-10y)[ 4x2 + 25y2 – 20xy + 4x2 + 25y2 + 20xy + 4x2 – 25y2]

= (-10y)[4x2 + 4x2 + 4x2 + 25y2]

= (-10y)[12x2 + 25y2}

= -120x2y – 250y3

Therefore, the result of (2x – 5y)3 – (2x + 5y)3 = -120x2y – 250y3

Q3. If a + b = 10 and ab = 21, Find the value of a3 + b3

Solution : Given here,

a + b = 10, ab = 21

Since, (a + b)3 = a3 + b3 + 3ab(a + b)        …………………….1

Put, a + b = 10 , ab = 21  in equation 1

=> (10)3 = a3 + b3 + 3(21)(10)

=>  1000 = a3 + b3 + 630

=> 1000 – 630 = a3 + b3

=> 370 = a3 + b3

Therefore, the value of  a3 + b3 = 370

Q4. If a – b = 4 and ab = 21, Find the value of a3 – b3

Solution: Given here,

a – b = 4 , ab = 21

Since, (a – b)3 = a3 – b3 – 3ab(a – b)        ——–  1

Put, a – b = 4 , ab = 21  in equation 1.

=> (4)3 = a3 – b3 – 3(21)(4)

=>  64  = a3 – b3 – 252

=> 64 + 252 = a3 – b3

=> 316 = a3 – b3

Therefore, the value of  a3 – b3 = 316

Q5. If ($x + \frac{1}{x}$) = 5 , Find the value of $x^{3} + \frac{1}{x^{3}}$

Solution: Given here, ($x + \frac{1}{x}$) = 5

Since, (a + b)3 = a3 + b3 + 3ab(a + b)          —– 1

Put, ($x + \frac{1}{x}$) = 5 in eq 1

($x + \frac{1}{x}$)3 = x3 + $\frac{1}{x^{3}}$ + 3(x * $\frac{1}{x}$)(x + $\frac{1}{x}$)

53    =  x3 + $\frac{1}{x^{3}}$ + 3(x * $\frac{1}{x}$)(x + $\frac{1}{x}$)

125 = x3 + $\frac{1}{x^{3}}$ + 3(x + $\frac{1}{x}$)

125 = x3 + $\frac{1}{x^{3}}$ + 3(5)

125 = x3 + $\frac{1}{x^{3}}$ + 15

125 – 15 = x3 + $\frac{1}{x^{3}}$

x3 +$\frac{1}{x^{3}}$ = 110

Therefore, the result is x3 + $\frac{1}{x^{3}}$ = 110

Q6. If ($x – \frac{1}{x}$) = 7 , Find the value of $x^{3} – \frac{1}{x^{3}}$

Solution:

Given here, ($x – \frac{1}{x}$) = 7

Since (a – b)3 = a3 – b3 – 3ab(a – b)        ——–  1

Put, ($x – \frac{1}{x}$) = 7, in equation 1.

($x – \frac{1}{x}$)3 = $x^{3} – \frac{1}{x^{3}}$ – 3(x * $\frac{1}{x}$)(x – $\frac{1}{x}$)

73 =  x$\frac{1}{x^{3}}$ – 3(x – $\frac{1}{x}$)

343   =  x3 –$\frac{1}{x^{3}}$ – (3 * 7)

343 = x3 –$\frac{1}{x^{3}}$ – 21

343 + 21 = x3 –$\frac{1}{x^{3}}$

x3 –$\frac{1}{x^{3}}$ = 364

Therefore, the result is x3 –$\frac{1}{x^{3}}$ = 364

Q7. If ($x – \frac{1}{x}$) = 5 , Find the value of $x^{3} – \frac{1}{x^{3}}$

Solution: Given here, ($x – \frac{1}{x}$) = 5

Since, (a – b)3 = a3 – b3 – 3ab(a – b)        ——–  1

Put ($x – \frac{1}{x}$) = 5, in equation 1

($x – \frac{1}{x}$)3 = x3 – $\frac{1}{x^{3}}$ – 3(x * $\frac{1}{x}$)(x – $\frac{1}{x}$)

53 =  x$\frac{1}{x^{3}}$ – 3(x – $\frac{1}{x}$)

125 =  x3 – $\frac{1}{x^{3}}$ – (3 * 5)

125 = x3 – $\frac{1}{x^{3}}$ – 15

125 + 15 = x3 – $\frac{1}{x^{3}}$

x3 – $\frac{1}{x^{3}}$ = 140

Therefore, the result is x3 – $\frac{1}{x^{3}}$ = 140

Q8. If ($x^{2} + \frac{1}{x^{2}}$) = 51 , Find the value of $x^{3} – \frac{1}{x^{3}}$

Solution: Given here,  ($x^{2} + \frac{1}{x^{2}}$) = 51

Since,  ( x – y)2 = x2 + y2 – 2xy                ——– 1

Put ($x^{2} + \frac{1}{x^{2}}$) = 51, in eq 1

(x – $\frac{1}{x}$)2 = x2 + $\frac{1}{x^{2}}$ – 2 * x * $\frac{1}{x}$

(x – $\frac{1}{x}$) = x2 + $\frac{1}{x^{2}}$ – 2

(x – $\frac{1}{x}$) = 51 – 2

(x – $\frac{1}{x}$) = 49

(x – $\frac{1}{x}$)  = $\sqrt{49}$

(x – $\frac{1}{x}$)  = ±7

We have to find $x^{3} – \frac{1}{x^{3}}$

Since, a3 – b3 = (a – b)(a2 + b2 + ab)

$x^{3} – \frac{1}{x^{3}}$ = (x – $\frac{1}{x}$)(x^{2} + $\frac{1}{x^{2}}$ + (x * $\frac{1}{x}$)

We have here;

(x –$\frac{1}{x}$)  = 7 and ($x^{2} + \frac{1}{x^{2}}$) = 51

$x^{3} – \frac{1}{x^{3}}$ = 7(51 + 1)

$x^{3} – \frac{1}{x^{3}}$ = 7(52)

$x^{3} – \frac{1}{x^{3}}$ = 364

Therefore, the value of  $x^{3} – \frac{1}{x^{3}}$ = 364

Q9.  If ($x^{2} + \frac{1}{x^{2}}$) = 98 , Find the value of $x^{3} + \frac{1}{x^{3}}$

Solution: Given here,  ($x^{2} + \frac{1}{x^{2}}$) = 98

Since , ( x + y)2 = x2 + y2 + 2xy                ——– 1

Put ($x^{2} + \frac{1}{x^{2}}$) = 98, in eq 1

(x + $\frac{1}{x}$)2 = x2 + $\frac{1}{x^{2}}$ + 2 * x * $\frac{1}{x}$

(x + $\frac{1}{x}$) = x2 + $\frac{1}{x^{2}}$ + 2

(x + $\frac{1}{x}$) = 98 + 2

(x + $\frac{1}{x}$) = 100

(x + $\frac{1}{x}$)  = $\sqrt{100}$

(x + $\frac{1}{x}$)  = ±10

We have to find $x^{3} + \frac{1}{x^{3}}$

Since, a3 + b3 = (a + b)(a2 + b2 – ab)

$x^{3} + \frac{1}{x^{3}}$ = (x + $\frac{1}{x}$)(x$\frac{1}{x^{2}}$ – (x * $\frac{1}{x}$)

(x + $\frac{1}{x}$)  = 10 and ($x^{2} + \frac{1}{x^{2}}$) = 98

$x^{3} + \frac{1}{x^{3}}$ = 10(98 – 1)

$x^{3} + \frac{1}{x^{3}}$ = 10(97)

$x^{3} + \frac{1}{x^{3}}$ = 970

Therefore, the value of  $x^{3} + \frac{1}{x^{3}}$ = 970

Q10. If 2x + 3y = 13 and xy = 6, Find the value of 8x3 + 27y3

Solution: Given here , 2x + 3y = 13 , xy = 6

Taking cube of both the sides for 2x + 3y = 13, we get;

(2x + 3y)3 = 133

Since, (a + b)3 = a3 + b3 + 3ab(a + b), thus,

=> 8x3 + 27y3 + 3(2x)(3y)(2x + 3y) = 2197

=> 8x3 + 27y3 + 18xy(2x + 3y) = 2197

Put, 2x + 3y = 13, xy = 6

=>  8x3 + 27y3 + 18(6)(13) = 2197

=>  8x3 + 27y3 + 1404 = 2197

=>  8x3 + 27y3  = 2197 – 1404

=>  8x3 + 27y3  = 793

Therefore, the value of  8x3 + 27y3  = 793

Q11. If  3x – 2y = 11 and xy = 12, Find the value of 27x3 – 8y3

Solution : Given here,  3x – 2y = 11 , xy = 12

Since, (a – b)3 = a3 – b3 – 3ab(a + b)

(3x – 2y)3 = 113

=> 27x3 – 8y3 – (18 * 12 * 11) = 1331

=> 27x3 – 8y3 – 2376  = 1331

=> 27x3 – 8y3  = 1331 + 2376

=> 27x3 – 8y3  = 3707

Therefore, the value of 27x3 – 8y3  = 3707

Q12. If $x^{4} + (\frac{1}{x^{4}})$ = 119 , Find the value of $x^{3} – (\frac{1}{x^{3}})$

Solution: Given here, $x^{4} + (\frac{1}{x^{4}})$ = 119           ——– 1

Since, (x + y)2 = x2 + y2 + 2xy

Now, put $x^{4} + (\frac{1}{x^{4}})$ = 119 in eq 1

($x^{2} + (\frac{1}{x^{2}})$)2 = x4 + ($\frac{1}{x^{4}}$) + (2*x2* $\frac{1}{x^{2}}$)

= x4 + ($\frac{1}{x^{4}}$) + 2

= 119 + 2

= 121

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

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

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

Now, let us find (x – $\frac{1}{x}$)

Since, (x – y)2 = x2 + y2 – 2xy

(x – $\frac{1}{x}$)2 = x2 + $\frac{1}{x^{2}}$ – (2*x*$\frac{1}{x}$

= x2 + $\frac{1}{x^{2}}$ – 2

= 11- 2

= 9

(x – $\frac{1}{x}$) = $\sqrt{9}$

= ±3

We have to find $x^{3} – (\frac{1}{x^{3}})$

Since, a3 – b3 = (a – b)(a2 + b2 – ab)

$x^{3} – (\frac{1}{x^{3}})$ = (x – $\frac{1}{x}$)( $x^{2} + (\frac{1}{x^{2}})$ + x * $\frac{1}{x}$

Now we get here, $x^{2} + (\frac{1}{x^{2}})$  = 11 and (x – $\frac{1}{x}$) = 3

$x^{3} – (\frac{1}{x^{3}})$ = 3(11 + 1)

= 3(12)

= 36

Therefore, the value of $x^{3} – (\frac{1}{x^{3}})$ = 36

Q13. Evaluate each of the following

(a) (103)3

(b) (98)3

(c) (9.9)3

(d) (10.4)3

(e) (598)3

(f) (99)3

Solution: Given here,

(a) (103)3

Since,  (a + b)3 = a3 + b3 + 3ab(a + b)

=> (103)3 can be written as (100 + 3)3

So, a = 100 and b = 3;

(103)3 =  (100 + 3)3

= (100)3 + (3)3 + 3(100)(3)(100 + 3)

= 1000000 + 27 + (900*103)

= 1000000 + 27 + 92700

= 1092727

Therefore, the value of (103)3 = 1092727

(b) (98)3

Since,  (a – b)3 = a3 – b3 – 3ab(a – b)

=> (98)3 can be written as (100 – 2)3

So, a = 100 and b = 2

(98)3 =  (100 – 2)3

= (100)3 – (2)3 – 3(100)(2)(100 – 2)

= 1000000 – 8 – (600*102)

= 1000000 – 8 – 58800

= 941192

Therefore, the value of (98)3 = 941192

(c) (9.9)3

Since, (a – b)3 = a3 – b3 – 3ab(a – b)

=> (9.9)3 can be written as (10 – 0.1)3

So, a = 10 and b = 0.1

(9.9)3 =  (10 – 0.1)3

= (10)3 – (0.1)3 – 3(10)(0.1)(10 – 0.1)

= 1000 – 0.001 – (3*9.9)

= 1000 – 0.001 – 29.7

= 1000 – 29.701

= 970.299

Therefore, the value of (9.9)3 = 970.299

(d) (10.4)3

Since, (a + b)3 = a3 + b3 + 3ab(a + b)

=> (10.4)3 can be written as (10 + 0.4)3

So, a = 10 and b = 0.4

(10.4)3 =  (10 + 0.4)3

= (10)3 + (0.4)3 + 3(10)(0.4)(10 + 0.4)

= 1000 + 0.064 + (12*10.4)

= 1000 + 0.064 + 124.8

= 1000 + 124.864

= 1124.864

Therefore, the value of (10.4)3 = 1124.864

(e) (598)3

Since,  (a – b)3 = a3 – b3 – 3ab(a – b)

=> (598)3 can be written as (600 – 2)3

So, a = 600 and b = 2

(598)3 =  (600 – 2)3

= (600)3 – (2)3 – 3(600)(2)(600 – 2)

= 216000000 – 8 – (3600*598)

= 216000000 – 8 – 2152800

= 216000000 – 2152808

= 213847192

Therefore, the value of (598)3 = 213847192

(f) (99)3

Since, we know, (a – b)3 = a3 – b3 – 3ab(a – b)

=> (99)3 can be written as (100 – 1)3

So, a = 100 and b = 1

(99)3      =  (100 – 1)3

= (100)3 – (1)3 – 3(100)(1)(100 – 1)

= 1000000 – 1 – (300*99)

= 1000000 – 1 – 29700

= 1000000 – 29701

= 970299

Therefore, the value of (99)3 = 970299

Q14. Evaluate each of the following

(a) 1113 – 893

(b) 463 + 343

(c) 1043 + 963

(d) 933 – 1073

Solution: Given here,

(a) 1113 – 893

The given expression can be written as; (100 + 11)3 – (100 – 11)3

Since, (a + b)3 – (a – b)3 = 2[b3 + 3ab2]

So, here, a= 100 b = 11.

Now,

(100 + 11)3 – (100 – 11)3      = 2[113 + 3(100)2(11)]

= 2[1331 + 330000]

= 2[331331]

= 662662

Therefore, the value of  1113 – 893 = 662662

(b) 463 + 343

The given expression can be written as; (40 + 6)3 + (40 – 6)3

Since, (a + b)3 + (a – b)3 = 2[a3 + 3ab2]

So, a= 40 , b = 4

(40 + 6)3 + (40 – 6)3      = 2[403 + 3(6)2(40)]

= 2[64000 + 4320]

= 2[68320]

= 1366340

Therefore, the value of  463 + 343 = 1366340

(c) 1043 + 963

The given expression can be written as; (100 + 4)3 + (100 – 4)3

Since, (a + b)3 + (a – b)3 = 2[a3 + 3ab2]

So, a= 100 b = 4

(100 + 4)3 – (100 – 4)3      = 2[1003 + 3(4)2(100)]

= 2[1000000 + 4800]

= 2[1004800]

= 2009600

Therefore, the value of  1043 + 963 = 2009600

(d) 933 – 1073

The given expression can be written as; (100 – 7)3 – (100 + 7)3

Since, (a – b)3 – (a + b)3 = -2[b3 + 3ba2]

So, a= 93,  b = 107

(100 – 7)3 – (100 + 7)3      = -2[73 + 3(100)2(7)]

= -2[343 + 210000]

= -2[210343]

= -420686

Therefore, the value of  933 – 1073 = -420686

Q15. If x + $\frac{1}{x}$ = 3, calculate $x^{2} + \frac{1}{x^{2}}$, $x^{3} + \frac{1}{x^{3}}$, $x^{4} + \frac{1}{x^{4}}$

Solution: Given here, x + $\frac{1}{x}$ = 3

Since, (x + y)2 = x2 + y2 + 2xy

(x + $\frac{1}{x}$)2 = x2 + $\frac{1}{x^{2}} + (2*x*\frac{1}{x})$

32 = x2 + $\frac{1}{x^{2}}$ + 2

9 – 2 = x2 + $\frac{1}{x^{2}}$

x2 + $\frac{1}{x^{2}}$ = 7

On squaring both the sides, we get;

( x2 + $\frac{1}{x^{2}}$)2 = 72

x4 + $\frac{1}{x^{4}}$ + 2 * x2 * $\frac{1}{x^{2}}$ = 49

x4 + $\frac{1}{x^{4}}$ + 2 = 49

x4 + $\frac{1}{x^{4}}$        = 49 – 2

x4 + $\frac{1}{x^{4}}$        = 47

Again, taking cube on both sides, we get;

(x + $\frac{1}{x}$)3 = 33

x3 + $\frac{1}{x^{3}}$ + 3x*$\frac{1}{x}$(x + $\frac{1}{x})$ = 27

x3 + $\frac{1}{x^{3}}$ + (3*3) = 27

x3 + $\frac{1}{x^{3}}$ + 9 = 27

x3 + $\frac{1}{x^{3}}$         = 27 – 9

x3 + $\frac{1}{x^{3}}$        = 18

Therefore, the values are x2 + $\frac{1}{x^{2}}$ = 7, x4 + $\frac{1}{x^{4}}$ = 47, x3 + $\frac{1}{x^{3}}$  = 18

Q16. If x4 + $\frac{1}{x^{4}}$ = 194, calculate $x^{2} + \frac{1}{x^{2}}$, $x^{3} + \frac{1}{x^{3}}$, $x + \frac{1}{x}$

Solution: Given here,

x4 + $\frac{1}{x^{4}}$ = 194          ——— 1

By adding and subtracting (2*$x^{2}* \frac{1}{x^{2}}$) on left side in the above given equation, we get;

x4 + $\frac{1}{x^{4}}$ + (2*$x^{2}* \frac{1}{x^{2}}$) – 2(2*$x^{2}* \frac{1}{x^{2}}$)                         =  194

x4 + $\frac{1}{x^{4}}$ + (2*$x^{2}* \frac{1}{x^{2}}$) – 2 = 194

(x2 + $\frac{1}{x^{2}}$)2 – 2    =  194

(x2 + $\frac{1}{x^{2}}$)2      = 194 + 2

(x2 + $\frac{1}{x^{2}}$)2 = 196

(x2 + $\frac{1}{x^{2}}$) = $\sqrt{196}$

(x2 + $\frac{1}{x^{2}}$) = 14                ———– 2

Now, by adding and subtracting (2*x* $\frac{1}{x}$) on left side in eq 2, we get;

(x2 + $\frac{1}{x^{2}}$) + (2*x* $\frac{1}{x}$) – (2*x* $\frac{1}{x}$) = 14

(x + $\frac{1}{x}$)2 – 2 = 14

(x + $\frac{1}{x}$)2  = 14 + 2

(x + $\frac{1}{x}$)2  = 16

(x + $\frac{1}{x}$) = $\sqrt{16}$

(x + $\frac{1}{x}$) = 4         ———- 3

Now, taking cube of eq 3 on both the sides, we get;

(x + $\frac{1}{x}$)3    = 43

Since, (a + b)3 = a3 + b3 + 3ab(a + b)

x3 + $\frac{1}{x^{3}}$ + 3*x*$\frac{1}{x}$(x + $\frac{1}{x}$) = 64

x3 + $\frac{1}{x^{3}}$ + (3*4) = 64

x3 + $\frac{1}{x^{3}}$ = 64 – 12

x3 + $\frac{1}{x^{3}}$ = 52

Therefore, the values of (x2 + $\frac{1}{x^{2}}$)2 = 196, (x + $\frac{1}{x}$) = 4 , x3 + $\frac{1}{x^{3}}$ = 52

Q17. Find the values of 27x3 + 8y3 , if

(a) 3x + 2y = 14 and xy = 8

(b) 3x + 2y = 20 and xy = $\frac{14}{9}$

Solution:

(a) Given,  3x + 2y = 14 and xy = 8

On taking cube both the sides, we get;

(3x + 2y)3 = 143

Since, (a + b)3 = a3 + b3 + 3ab(a + b)

27x3 + 8y3 + 3(3x)(2y)(3x + 2y) = 2744

27x3 + 8y3 + 18xy(3x + 2y) = 2744

27x3 + 8y3 + 18(8)(14) = 2744

27x3 + 8y3 + 2016 = 2744

27x3 + 8y3  =  2744 – 2016

27x3 + 8y3  =  728

Therefore, the value of  27x3 + 8y3  =  728

(b) Given here, 3x + 2y = 20 and xy = 14/9

On taking cube both the sides, we get;

(3x + 2y)3 = 203

Since, (a + b)3 = a3 + b3 + 3ab(a + b)

27x3 + 8y3 + 3(3x)(2y)(3x + 2y) = 8000

27x3 + 8y3 + 18xy(3x + 2y) = 8000

27x3 + 8y3 + 18($\frac{14}{9}$)(20) = 8000

27x3 + 8y3 + 560 = 8000

27x3 + 8y3         = 8000 – 560

27x3 + 8y3         = 7440

Therefore, the value of 27x3 + 8y3 = 7440

Q18. Find the value of 64x3 – 125z3 , if 4x – 5z = 16 and xz = 12

Solution: Given here, 64x3 – 125z3

And, 4x – 5z = 16 and xz = 12

On taking cube for 4x – 5z = 16, both the sides, we get;

(4x – 5z)3 = 163

Since, (a – b)3 = a3 – b3 – 3ab(a – b)

(4x)3 – (5z)3 – 3(4x)(5z)(4x – 5z) = 163

64x3 – 125z3 – 60(xz)(16) = 4096

64x3 – 125z3 – 60(12)(16) = 4096

64x3 – 125z3 – 11520 = 4096

64x3 – 125z3  = 4096 + 11520

64x3 – 125z3  = 15616

Therefore, the value of  64x3 – 125z3  = 15616

Q19. If x – $\frac{1}{x}$ = 3 + 2$\sqrt{2}$ , Find the value of x3$\frac{1}{x^{3}}$

Solution: Given here, x – $\frac{1}{x}$ = 3 + 2$\sqrt{2}$

On taking cube of x – $\frac{1}{x}$ = 3 + 2$\sqrt{2}$ on both sides, we get;

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

Since, (a + b)3 = a3 + b3 + 3ab(a + b)

x3$\frac{1}{x^{3}}$ – 3*x*$\frac{1}{x}$(x – $\frac{1}{x}$) = 32 + (2$\sqrt{2}$)3+ 3*3*2$\sqrt{2}$(3 + 2$\sqrt{2}$)

x3$\frac{1}{x^{3}}$ – 3(3 + 2$\sqrt{2}$) = 27 + 16$\sqrt{2}$ + 18$\sqrt{2}$(3 + 2$\sqrt{2}$)

x3$\frac{1}{x^{3}}$ = 27 + 16$\sqrt{2}$ + 54$\sqrt{2}$ + 72 + 9 + 6$\sqrt{2}$

x3$\frac{1}{x^{3}}$ = 108 + 76 $\sqrt{2}$

Therefore, the value of x3$\frac{1}{x^{3}}$ = 108 + 76 $\sqrt{2}$