Question

a3-b3 =(a-b)(a2+b2+ab)

Answer 1

HELLO! !! ....


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●From the identity of cube of a binomial, we have

(a + b)3 = a3 + b3 + 3ab(a + b)

∵ a3 + b3 = (a + b)3 - 3ab(a + b)

●Taking (a + b) as common,

= (a + b)[(a + b)2 - 3ab]

= (a + b)(a2 + b2 + 2ab - 3ab)

= (a + b)(a2 + b2 - ab)

=> Hence a3 + b3= (a + b)(a2 + b2 - ab)

●From the identity of cube of a binomial, we have

(a - b)3 = a3 - b3 - 3ab(a - b)

∵ a3 - b3 = (a - b)3 + 3ab(a - b)

●Taking (a - b) as common,

= (a - b)[(a - b)2 + 3ab]

= (a - b)(a2 + b2 - 2ab + 3ab)

= (a - b)(a2 + b2 + ab)

=> Hence a3 - b3= (a - b)(a2 + b2 + ab)


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Hope it helps u !!!!

# Nikky

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