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L.H.S. = (a + b) 3 + (b + c) 3 + (c + a) 3 - 3(a + b)(b + c)(c + a) = {(a + b) + (b + c) + (c + a)} [(a + b) 2 + (b + c) 2 + (c + a) 2 - (a + b)(b + c) - (b + c)(c + a) - (c + a)(a + b)] | Using Identity VIII = 2(a + b + c) [a 2 + 2ab + b 2 + b 2 + 2bc + c 2 + c 2 + 2ca + c 2 - ab - ac - b 2 - bc - bc - ba - c 2 - ca - ca - cb - a 2 - ab] | Using Identity I = 2(a + b + c)(a 2 + b 2 + c 2 - ab - be - ca) = 2(a 3 + b 3 + c 3 - 3abc). | Using Identity VIII

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Polynomials

Short Answer Type

161.

Simplify: (x + y)³ - (x - y)³ - 6y(x²- y²)

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162. If x² + px + q = (x + a)(x + b), then factorise x² + pxy + qy².

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163. Prove that: (a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a) = 2(a³ + b³+ c³ - 3abc).

L.H.S. = (a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a)
= {(a + b) + (b + c) + (c + a)} [(a + b)² + (b + c)² + (c + a)² - (a + b)(b + c) - (b + c)(c + a) - (c + a)(a + b)]
| Using Identity VIII
= 2(a + b + c) [a² + 2ab + b² + b² + 2bc + c² + c² + 2ca + c² - ab - ac - b² - bc - bc - ba - c² - ca - ca - cb - a² - ab]
| Using Identity I
= 2(a + b + c)(a² + b² + c² - ab - be - ca)
= 2(a³ + b³ + c³ - 3abc).
| Using Identity VIII