Question

Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a).

Answer 1

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.