# RD Sharma Solutions Class 9 Algebraic Identities Exercise 4.4

## RD Sharma Solutions Class 9 Chapter 4 Ex 4.4

Q1. Find the following products

(a) (3x + 2y)(9x2 – 6xy + 4y2)

(b) (4x – 5y)(16x2 + 20xy + 25y2)

(c) (7p4 + q)(49p8 – 7p4q + q2)

(d) ($\frac{x}{2}$ + 2y)($\frac{x^{2}}{4}$ – xy + 4y2)

(e) $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

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

(g) $(\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)$

(h) $(\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)$

(i) (1 – x)(1 + x + x^{2})

(j) (1 + x)(1 – x + x^{2})

(k) $(x^{2} – 1)(x^{4} + x^{2} + 1)$

(l) $(x^{2} + 1)(x^{6} – x^{3} + 1)$

Sol :

(a) (3x + 2y)(9x2 – 6xy + 4y2)

As given, (3x + 2y)(9x2 – 6xy + 4y2)

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

So, (3x + 2y)(9x2 – 6xy + 4y2) can be solved as

=> (3x + 2y)[(3x)2 – (3x)(2y) + (2y)2)]

=> (3x)3 + (2y)3

=> 27x3 + 8y3

Therefore, the product value of (3x + 2y)(9x2 – 6xy + 4y2) is  27x3 + 8y3

(b) (4x – 5y)(16x2 + 20xy + 25y2)

As Given, (4x – 5y)(16x2 + 20xy + 25y2)

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

(4x – 5y)(16x2 + 20xy + 25y2) can be solved as

=> (4x – 5y)[(4x)2 + (4x)(5y) + (5y)2)]

=> (4x)3 – (5y)3

=> 16x3 – 25y3

Therefore, the product value of (4x – 5y)(16x2 + 20xy + 25y2) = 16x3 – 25y3

(c) (7p4 + q)(49p8 – 7p4q + q2)

As Given, (7p4 + q)(49p8 – 7p4q + q2)

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

(7p4 + q)(49p8 – 7p4q + q2) can be solved as

=> (7p4 + q)[(7p4)2 – (7p4)(q) + (q)2)]

=> (7p4)3 + (q)3

=> 343p12 + q3

Therefore, the product value of (7p4 + q)(49p8 – 7p4q + q2) = 343p12 + q3

(d) ($\frac{x}{2}$ + 2y)( $\frac{x^{2}}{4}$ – xy + 4y2)

Sol :

As Given, ($\frac{x}{2}$ + 2y)( $\frac{x^{2}}{4}$ – xy + 4y2)

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

($\frac{x}{2}$ + 2y)( $\frac{x^{2}}{4}$ – xy + 4y2) can be solved as

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

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

=> $\frac{x^{3}}{8} + 8y^{3}$

(e) $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

Sol:

As Given, $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

$(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

can be solved as ,

=> $(\frac{3}{x} – \frac{5}{y})(\frac{3}{x}$)2 +($\frac{5}{y}$)2 + $(\frac{3}{x})( \frac{5}{y})$

=> $(\frac{3}{x})^{3} – (\frac{5}{y})^{3}$

=> $(\frac{27}{x^{3}}) – (\frac{125}{y^{3}})$

Therefore, the product value of$(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$ = $(\frac{27}{x^{3}}) – (\frac{125}{y^{3}})$

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

Sol:

As Given, $(3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})$

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

$(3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})$ can be solved as,

=> $(3 + \frac{5}{x})[(3^{2}) – 3(\frac{5}{x}) + (\frac{5}{x})^{2}]$

=> $(3)^{3} + (\frac{5}{x})^{3}$

=> 27 + $\frac{125}{x^{3}}$

Therefore, the product value of $(3 + \frac{5}{x})(9 – \frac{15}{x} + \frac{25}{x^{2}})$ is 27 + $\frac{125}{x^{3}}$

(g) $(\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)$

Sol:

As Given, $(\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)$

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

$(\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)$ can be solved as,

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

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

=> $\frac{8}{x^{3}} + 9x^{3}$

Therefore, the product value of $(\frac{2}{x} + 3x)(\frac{4}{x^{2}} + 9x^{2} – 6)$ is $\frac{8}{x^{3}} + 9x^{3}$

(h) $(\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)$

Sol:

As Given, $(\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)$

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

$(\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)$ can be solved as,

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

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

=> $(\frac{3}{x})^{3} – (2x^{2})^{3}$

=> $\frac{27}{x^{3}} – 8x^{6}$

Therefore, the product value of $(\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)$ is $\frac{27}{x^{3}} – 8x^{6}$

(i) (1 – x)(1 + x + x^{2})

Sol:

As Given, (1 – x)(1 + x + x^{2})

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

(1 – x)(1 + x + x^{2}) can be solved as,

=> (1 – x)[(12 + (1)(x)+ x2)]

=> (1)3 – (x)3

=> 1 – x3

Therefore, the product value of (1 – x)(1 + x + x^{2}) is 1 – x3

(j) (1 + x)(1 – x + x^{2})

Sol:

As Given, (1 + x)(1 – x + x^{2})

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

(1 + x)(1 – x + x^{2}) can be solved as,

=> (1 + x)[(12 – (1)(x) + x2)]

=> (1)3 + (x)3

=> 1 + x3

Therefore, the product value of (1 + x)(1 + x – x^{2}) is 1 + x3

(k) $(x^{2} – 1)(x^{4} + x^{2} + 1)$

Sol:

As Given, $(x^{2} – 1)(x^{4} + x^{2} + 1)$

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

$(x^{2} – 1)(x^{4} + x^{2} + 1)$ can be solved as,

=> (x2 – 1)[(x2)2 – 12 + (x2)(1)]

=> (x2)3 – 13

=> x6 – 1

Therefore, the product value of $(x^{2} – 1)(x^{4} + x^{2} + 1)$ is x6 – 1

(l) $(x^{2} + 1)(x^{6} – x^{3} + 1)$

Sol:

As Given, $(x^{2} + 1)(x^{6} – x^{3} + 1)$

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)

$(x^{2} + 1)(x^{6} – x^{3} + 1)$can be solved as,

=> (x3 + 1)[(x3)2 – (x3)(1) + 12]

=> (x3)3 + 13

=> x9 + 1

Therefore, the product value of $(x^{2} + 1)(x^{6} – x^{3} + 1)$ is x9 + 1

Q2. Find x= 3 and y = -1, Find the values of each of the following using in identity:

(a) (9x2 – 4x2)(81y4 + 36x2y2 + 16x4)

(b) $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

(c) $(\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})$

(d) $(\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})$

(e) $(\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})$

Sol:

(a) (9x2 – 4x2)(81y4 + 36x2y2 + 16x4)

As we know, a3 – b3 = (a – b)(a2 + b2 + ab)

Using the above identity for the product of (9x2 – 4x2)(81y4 + 36x2y2 + 16x4) , we get

=> (9x2 – 4x2)[(9y2)2 + (9)(4)x2y2 + (4x2)2]

=> (9y2)3 – (4x2)3

=> 729y6 – 64x6

Substitute the value x = 3, y = -1 in 729y6 – 64x6 we get,

=> 729y6 – 64x6

=> 729(-1)6 – 64(3)6

=> 729(1) – 64(729)

=> 729 – 46656

=> -45927

Therefore, the product value of (9x2 – 4x2)(81y4 + 36x2y2 + 16x4) = -45927

(b) $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

Sol:

Given,

As we know, a3 – b3 = (a – b)(a2 + b2 + ab)

Using the above identity for the product of $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$

we get,

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

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

=> $(\frac{27}{x^{3}}) – (\frac{x^{3}}{27})$ —– 1

Substitute x = 3 in eq 1

=> $(\frac{27}{3^{3}}) – (\frac{3^{3}}{27})$

=> $(\frac{27}{27}) – (\frac{27}{27})$

=> 0

Therefore, the value of $(\frac{3}{x} – \frac{5}{y})(\frac{9}{x^{2}} + \frac{25}{y^{2}} + \frac{15}{xy})$ is 0

(c) $(\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})$

Sol:

Given,

As we know, a3 + b3 = (a + b)(a2 + b2 – ab)

Using the above identity for the product of $(\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})$

we get,

=> $(\frac{x}{7} + \frac{y}{3})[(\frac{x}{7})^{2} + (\frac{y}{3})^{2} – (\frac{x}{7})(\frac{y}{3})]$

=> $(\frac{x}{7})^{3} + (\frac{y}{3})^{3}$

=> $(\frac{x^{3}}{343}) + (\frac{y^{3}}{27})$ —– 1

Substitute x = 3, y = -1 in eq 1

=> $(\frac{3^{3}}{343}) + (\frac{(-1)^{3}}{27})$

=> $(\frac{27}{343}) – (\frac{1}{27})$

Taking least common multiple, we get

=> $\frac{27*27}{343*27} – \frac{1*343}{27*343}$

=> $\frac{729}{9261} – \frac{343}{9261}$

=> $\frac{729 – 343}{9261}$

=> $\frac{386}{9261}$

Therefore, the value of $(\frac{x}{7} + \frac{y}{3})(\frac{x^{2}}{49} + \frac{y^{2}}{9} – \frac{xy}{21})$ = $\frac{386}{9261}$

(d) $(\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} -+ \frac{xy}{21})$

Sol:

Given,

As we know, a3 – b3 = (a – b)(a2 + b2 + ab)

Using the above identity for the product of$(\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})$

we get,

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

=> $(\frac{x}{4})^{3} – (\frac{y}{3})^{3}$

=> $(\frac{x^{3}}{64}) – (\frac{y^{3}}{27})$ —– 1

Substitute x = 3, y = -1 in eq 1

=> $(\frac{3^{3}}{343}) – (\frac{(-1)^{3}}{27})$

=> $(\frac{27}{64}) + (\frac{1}{27})$

Taking least common multiple, we get

=> $\frac{27*27}{64*27} + \frac{1*64}{27*64}$

=> $\frac{729}{1728} + \frac{64}{1728}$

=> $\frac{729 + 64}{1728}$

=> $\frac{793}{9261}$

Therefore, the value of $(\frac{x}{4} – \frac{y}{3})(\frac{x^{2}}{16} + \frac{y^{2}}{9} + \frac{xy}{21})$ = $\frac{793}{1728}$

(e) $(\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})$

Sol:

As we know, a3 + b3 = (a + b)(a2 + b2 – ab)

Using the above identity for the product of $(\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})$ we get,

=> $(\frac{5}{x} + 5x)[(\frac{5}{x})^{2} + (5x)^{2} – (\frac{5}{x})(5x)]$

=> $(\frac{5}{x})^{3} + (5x)^{3}$

=> $\frac{125}{x^{3}} + 125x^{3}$ ——- 1

Substitute x = 3, in eq 1

=> $\frac{125}{3^{3}} + 125(3)^{3}$

=> $\frac{125}{27} + 125*27$

=> $\frac{125}{27} + 3375$

Taking least common multiple, we get

=> $\frac{125}{27} + \frac{3375*27}{27*1}$

=> $\frac{125}{27} + \frac{91125}{27}$

=> $\frac{125 + 91125}{27}$

=> $\frac{91250}{25}$

Therefore, the value of $(\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})$ is $\frac{91250}{25}$

Q3. If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2

Sol : It is given that  a + b = 10, ab = 16

And we know, (a + b)3 = a3 + b3 + 3ab(a + b)

=> a3 + b3 = (a + b)3 – 3ab(a + b)

Putting the given values in the above identity, we get

=> a3 + b3 = (10)3 – 3(16)(10)

=> a3 + b3 = 1000 – 480

=> a3 + b3 = 520

let’s substitute , a3 + b3 = 520, a + b = 10 in a3 + b3 = (a + b)(a2 + b2 – ab)

a3 + b3 = (a + b)(a2 + b2 – ab)

520 = 10(a2 + b2 – ab)

$\frac{520}{10}$ = (a2 + b2 – ab)

=> (a2 + b2 – ab) = 52

Now, let’s find (a2 + b2 + ab)

Add and subtract 2ab in a2 + b2 + ab

=> a2 + b2 + ab – 2ab + 2ab

=> (a + b)2 – ab

Substitute a + b = 10, ab

=> a2 + b2 + ab = 102 – 16

= 100 – 16

= 84

Therfore, the values of (a2 + b2 – ab) = 52 and (a2 + b2 + ab) = 84

Q4. If a + b = 8 and ab = 6, find the value of a3 + b3

Sol:

It is given that, a + b = 8 and ab = 6

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

=> a3 + b3 = (a + b)3 – 3ab(a + b)

=> a3 + b3 = (8)3 – 3(6)(8)

=> a3 + b3 = 512 – 144

=> a3 + b3 = 368

Therfore, the value of a3 + b3 is 368

Q5. If a – b = 6 and ab = 20, find the value of a3 – b3

Sol:

It is given, a – b = 6 and ab = 20

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

=> a3 – b3 = (a – b)3 + 3ab(a – b)

=> a3 – b3 = (6)3 + 3(20)(6)

=> a3 – b3 = 216 + 360

=> a3 – b3 = 576

Therefore, the value of a3 – b3 is 576

Q6. If x = -2 and y = 1, by using an identity find the value of the following:

(a) (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)

(b) $(\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)$

(c) $(5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})$

Sol:

It is given that,

(a) (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)

And we know , a3 – b3 = (a – b)(a2 + b2 + ab)

Using above identity, (4y2 – 9x2)(16y4 + 36x2y2 + 81x4) we get ,

=> (4y2 – 9x2)[(4x)2 + 4y2*9x2 + (9x2)2)

=> (4y2)3 – (9x2)3

=> 64y6 – 729x6 ———- 1

Substitute x = -2 and y = 1 in eq 1

=> 64y6 – 729x6

=> 64(1)6 – 729(-2)6

=> 64 – 729(64)

=> 64(1 – 729)

=> 64(-728)

=> -46592

Therefore, the value of (4y2 – 9x2)(16y4 + 36x2y2 + 81x4) is -46592

(b) $(\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)$

Given x = -2

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

$(\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)$ can be witten as,

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

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

=> $(\frac{8}{x^{3}}) – (\frac{x^{3}}{8})$ ——- 1

Substitute x = -2 in eq 1

=> $(\frac{8}{(-2)^{3}}) – (\frac{(-2)^{3}}{8})$

=> $(\frac{8}{-8}) – (\frac{-8}{8})$

=> -1 + 1

=> 0

Therefore, the value of $(\frac{2}{x} – \frac{x}{2})(\frac{4}{x^{2}} + \frac{x^{2}}{4} + 1)$ is 0

(c) $(5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})$

Sol:

As we know, a3 + b3 = (a + b)(a2 + b2 – ab)

Using the above identity for $(5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})$ we get,

=> (5y + $\frac{15}{y}$)[(5y)2 + ($\frac{15}{y}$)2 – (5y)( $\frac{15}{y}$)]

=> (5y)3 + ($\frac{15}{y}$)3

=> 125y3 + ($\frac{3375}{y^{3}}$) ——– 1

Substitute y = 1 in eq 1

=> 125(1)3 + ($\frac{3375}{(1)^{2}}$)

=> 125 + 3375

=> 3500

Therefore, the value of $(5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})$ is 3500.

EXERCISE 4.5

Q1. Find the following products:

(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)

(b) (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)

(c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)

(d) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)

Sol:

It is given,

(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)

And we know that,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

Using the above identity,

(3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) = (3x)3 + (2y)3 + (2z)3 – 3(3x)(2y)(2z)

= 27x3 + 8y3 + 8z3 – 36xyz

Therfore, the value of (3x + 2y + 2z)(9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) is 27x3 + 8y3 + 8z3 – 36xyz

(b) (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)

As we know that,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

so, using the above identity

(4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx) = (4x)3 + (-3y)3 + (2z)3 -3(4x)(-3y)(2z)

= 64x3 – 27y3+ 8z3 + 72xyz

Hence, the value of (4x – 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx) is 64x3 – 27y3+ 8z3 + 72xyz

(c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)

As we know that,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

so, using the above identity

(2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) = (2a)3 + (-3b)3 + (-2c)3 – 3(2a)(-3b)(-2c)

= 8a3 – 27b3 – 8c3 – 36abc

Therfore, the value of (c) (2a – 3b – 2c)(4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) is 8a3 – 27b3 – 8c3 – 36abc

(d) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)

As  we know that,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

so, using the above identity

(3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz) = (3x)3 + (-4y)3 + (5z)3 -3(3x)(-4y)(5z)

= 27x3 – 64y3 + 125z3 + 180xyz

Therefore, the value of (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz) is 27x3 – 64y3 + 125z3 + 180xyz

Q2. If x + y + z = 8 and xy + yz + zx = 20, Find the value of x3 + y3 + z3 – 3xyz

Sol:
given, x + y + z = 8 and xy + yz + zx = 20

As we know that,

(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)

substituting the given values in the above identity

(x + y + z)2 = x2 + y2 + z2 + 2(20)

(x + y + z)2 = x2 + y2 + z2 + 40

82 = x2 + y2 + z2 + 40

64 – 40 = x2 + y2 + z2

x2 + y2 + z2 = 24

Now we know that,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

x3 + y3 + z3 – 3xyz = (x + y + z)[(x2 + y2 + z2) – (xy + yz + zx)]

As, x + y + z = 8, xy + yz + zx = 20, x2 + y2 + z2 = 24

x3 + y3 + z3 – 3xyz = 8[(24 – 20)]

= 8 * 4

= 32

Therefore, the value of x3 + y3 + z3 – 3xyz is 32

Q3. If a + b + c = 9 and ab + bc + ca = 26, Find the value of a3 + b3 + c3 – 3abc

Sol :

As it is given, a + b + c = 9 and ab + bc + ca = 26

As we know that,

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

(a + b + c)2 = a2 + b2 + c2 + 2(26)

(a + b + c)2 = a2 + b2 + c2 + 52

92 = a2 + b2 + c2 + 52

81 – 52 = a2 + b2 + c2

a2 + b2 + c2 = 29

Now,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 – 3abc = (a + b + c)[(a2 + b2 + c2) – (ab + bc + ca)]

as given , a + b + c = 9, ab + bc + ca = 26, a2 + b2 + c2 = 29

a3 + b3 + c3 – 3abc = 9[(29 – 26)]

= 9 * 3

= 27

Therefore, the value of a3 + b3 + c3 – 3abc is 27

Q4. If a + b + c = 9 and a2 + b2 + c2 = 35, Find the value of a3 + b3 + c3 – 3abc

Sol:

As given, a + b + c = 9 and a2 + b2 + c2 = 35

And we know that,

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

92 = 35 + 2(ab + bc + ca)

81 = 35 + 2(ab + bc + ca)

81 – 35 = 2(ab + bc + ca)

$\frac{46}{2}$ = ab + bc + ca

ab + bc + ca = 23

So we know that,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 – 3abc = (a + b + c)[(a2 + b2 + c2) – (ab + bc + ca)]

As given, a + b + c = 9, ab + bc + ca = 23, a2 + b2 + c2 = 35

a3 + b3 + c3 – 3abc = 9[(35 – 23)]

= 9 * 12

= 108

Therefore, the value of a3 + b3 + c3 – 3abc is 108

Q5. Evaluate:

(a) 253 – 753 + 503

(b) 483 – 303 – 183

(c) $(\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}$

(d) (0.2)3 – (0.3)3 + (0.1)3

Sol:

As given,

(a) 253 – 753 + 503

And we know that,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Given, a = 25, b = -75, c = 50

Substituting the given values of a, b ,c in the identity , we get

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = (25 – 75 + 50)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = 3abc

253 + (-75)3 + 503 = 3abc

= 3(25)(-75)(50)

= -281250

Therefore, the value 253 + (-75)3 + 503 = -281250

(b) 483 – 303 – 183

As we know that,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Given , a = 48, b = -30, c = -18

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

Substituting the given values of a, b ,c in the identity , we get

a3 + b3 + c3 = (48 – 30 – 18)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = 3abc

483 + (-30)3 + (-18)3 = 3abc

= 3(48)(-30)(-18)

= 77760

Therefore, the value 483 + (-30)3 + (-18)3 = 77760

(c) $(\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}$

As we know that,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

So, a= $\frac{1}{2}$, b= $\frac{1}{3}$, c= $\frac{-5}{6}$

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

Substituting the given values of a, b ,c in the identity , we get

a3 + b3 + c3 = ($\frac{1}{2}$ +$\frac{1}{3}$$\frac{5}{6}$)( a2 + b2 + c2 – ab – bc – ca) + 3abc

by applying least common multiple method

a3 + b3 + c3 = $(\frac{1*6}{2*6} + \frac{1*4}{3*4} – \frac{5*2}{6*2})$( a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = $(\frac{6}{12} + \frac{4}{12} – \frac{10}{12})$( a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = 0( a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = 3abc

($\frac{1}{2}$)3 + ($\frac{1}{3}$)3 – ($\frac{-5}{6}$)3 = 3*$\frac{1}{2}$* $\frac{1}{3}$* $\frac{-5}{6}$

= $\frac{1}{2}$ * $\frac{-5}{6}$

= $\frac{-5}{12}$

Therefore, the value of $(\frac{1}{2})^{3} + (\frac{1}{3})^{3} – (\frac{5}{6})^{3}$ is $\frac{-5}{12}$

(d) (0.2)3 – (0.3)3 + (0.1)3

As we know that,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Using the above identity, consider the value of a = 0.2, b = 0.3, c = 0.1

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

Substituting the values of a, b and c, we get

a3 + b3 + c3 = (0.2 – 0.3 + 0.1)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = (0)(a2 + b2 + c2 – ab – bc – ca) + 3abc

a3 + b3 + c3 = 3abc

(0.2)3 – (0.3)3 + (0.1)3 = 3abc

= 3(0.2)(-0.3)(0.1)

= -0.018

Therefore, the value (0.2)3 – (0.3)3 + (0.1)3 is 0.018