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