RD Sharma Solutions Class 9 Algebraic Identities Exercise 4.4

RD Sharma Solutions Class 9 Chapter 4 Exercise 4.4

RD Sharma Class 9 Solutions Chapter 4 Ex 4.4 Free Download

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)

Given, (3x + 2y)(9x2 – 6xy + 4y2)

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

(3x + 2y)(9x2 – 6xy + 4y2) can we written as

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

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

=> 27x3 + 8y3

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

(b) (4x – 5y)(16x2 + 20xy + 25y2)

Given, (4x – 5y)(16x2 + 20xy + 25y2)

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

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

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

=> (4x)3 – (5y)3

=> 16x3 – 25y3

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

(c) (7p4 + q)(49p8 – 7p4q + q2)

Given, (7p4 + q)(49p8 – 7p4q + q2)

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

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

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

=> 343p12 + q3

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

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

Sol :

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

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

(\(\frac{x}{2}\) + 2y)( \(\frac{x^{2}}{4}\) – xy + 4y2) can be written 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:

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

We know that, 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 written as,

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

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

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

Hence, the 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:

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

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

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

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

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

Hence, the 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:

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

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

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

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

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

Hence, the 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:

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

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

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

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

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

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

=> \(\frac{27}{x^{3}} – 8x^{6}\)

Hence, \((\frac{3}{x} – 2x^{2})(\frac{9}{x^{2}} + 4x^{4} – 6x)\) is \(\frac{27}{x^{3}} – 8x^{6}\)

(i) (1 – x)(1 + x + x^{2})

Sol:

Given, (1 – x)(1 + x + x^{2})

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

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

=> (1 – x)[(12 + (1)(x)+ x2)]

=> (1)3 – (x)3

=> 1 – x3

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

(j) (1 + x)(1 – x + x^{2})

Sol:

Given, (1 + x)(1 – x + x^{2})

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

=> (1 + x)[(12 – (1)(x) + x2)]

=> (1)3 + (x)3

=> 1 + x3

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

(k) \((x^{2} – 1)(x^{4} + x^{2} + 1)\)

Sol:

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

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

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

=> (x2 – 1)[(x2)2 – 12 + (x2)(1)]

=> (x2)3 – 13

=> x6 – 1

Hence, \((x^{2} – 1)(x^{4} + x^{2} + 1)\) is x6 – 1

(l) \((x^{2} + 1)(x^{6} – x^{3} + 1)\)

Sol:

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

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

=> (x3 + 1)[(x3)2 – (x3)(1) + 12]

=> (x3)3 + 13

=> x9 + 1

Hence, the 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)

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

(9x2 – 4x2)(81y4 + 36x2y2 + 16x4) can be written as,

=> (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

Hence, 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,

We know that, 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 written as,

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

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

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

Substitute x = 3 in eq 1

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

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

=> 0

Hence, 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,

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

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

Can be written as,

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

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

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

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

=> \(\frac{729}{9261} – \frac{343}{9261}\)

=> \(\frac{729 – 343}{9261}\)

=> \(\frac{386}{9261}\)

Hence, 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,

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

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

Can be written as,

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

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

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

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

=> \(\frac{729}{1728} + \frac{64}{1728}\)

=> \(\frac{729 + 64}{1728}\)

=> \(\frac{793}{9261}\)

Hence, 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:

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

\((\frac{5}{x} + 5x)(\frac{25}{x^{2}} – 25 + 25x^{2})\) can be written as,

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

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

=> \(\frac{125}{x^{3}} + 125x^{3}\) ——- 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

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

=> \(\frac{125}{27} + \frac{91125}{27}\)

=> \(\frac{125 + 91125}{27}\)

=> \(\frac{91250}{25}\)

Hence, 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 :

Given, a + b = 10, ab = 16

We know that, (a + b)3 = a3 + b3 + 3ab(a + b)

=> a3 + b3 = (a + b)3 – 3ab(a + b)

=> a3 + b3 = (10)3 – 3(16)(10)

=> a3 + b3 = 1000 – 480

=> a3 + b3 = 520

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, we need to 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

Hence, 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:

Given, a + b = 8 and ab = 6

We know that, 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

Hence, the value of a3 + b3 is 368

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

Sol:

Given, a – b = 6 and ab = 20

We know that, 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

Hence, 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:

Given,

(a) (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)

We know that, a3 – b3 = (a – b)(a2 + b2 + ab)

(4y2 – 9x2)(16y4 + 36x2y2 + 81x4) can be written as,

=> (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

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

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

here x = -2

We know that, 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

Hence, 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:

We know that, a3 + b3 = (a + b)(a2 + b2 – ab)

\((5y + \frac{15}{y})(25y^{2} – 75 + \frac{225}{y^{2}})\) can be written as,

=> (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

Hence, 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:

Given,

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

we know that,

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

so,

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

= 27x3 + 8y3 + 8z3 – 36xyz

Hence, 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)

we know that,

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

so,

(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)

we know that,

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

so,

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

= 8a3 – 27b3 – 8c3 – 36abc

Hence, 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)

we know that,

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

so,

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

= 27x3 – 64y3 + 125z3 + 180xyz

Hence, 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

We know that,

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

(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

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)]

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

x3 + y3 + z3 – 3xyz = 8[(24 – 20)]

= 8 * 4

= 32

Hence, 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 :

Given, a + b + c = 9 and ab + bc + ca = 26

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

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)]

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

a3 + b3 + c3 – 3abc = 9[(29 – 26)]

= 9 * 3

= 27

Hence, 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:

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

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

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)]

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

a3 + b3 + c3 – 3abc = 9[(35 – 23)]

= 9 * 12

= 108

Hence, 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:

Given,

(a) 253 – 753 + 503

we know that,

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

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

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

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

(b) 483 – 303 – 183

we know that,

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

here, 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

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

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

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

we know that,

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

here, 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

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

by using least common multiple

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*\(\frac{1}{2}\)* \(\frac{1}{3}\)* \(\frac{-5}{6}\)

= \(\frac{1}{2}\) * \(\frac{-5}{6}\)

= \(\frac{-5}{12}\)

Hence, 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

we know that,

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

here, 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

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

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


Practise This Question

Ram drew a line segment AB of length 3cm and another line segment CB of length 4cm which is perpendicular to line segment AB. What is the length of the third side of the triangle?