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प्रश्न
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
उत्तर
To prove: (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a)
L.H.S = [(a + b + c)3 – a3] – (b3 + c3)
= (a + b + c – a)[(a + b + c)2 + a2 + a(a + b + c)] – [(b + c)(b2 + c2 – bc)] ...[Using identity, a3 + b3 = (a + b)(a2 + b2 – ab) and a3 – b3 = (a – b)(a2 + b2 + ab)]
= (b + c)[a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac] – (b + c)(b2 + c2 – bc)
= (b + c)[b2 + c2 + 3a2 + 3ab + 3ac – b2 – c2 + 3bc]
= (b + c)[3(a2 + ab + ac + bc)]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)[(a + c)(a + b)]
= 3(a + b)(b + c)(c + a) = R.H.S
Hence proved.
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