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

1.	Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
Answer» We have, L.H.S. = (a + b + c)³ – a³ -b³ – c³ = {(a + b + c)³ – (a)³} – (b³ + c³) = (a + b + c – a){(a + b + c)² + a(a + b + c) + a²} – (b + c){b² – bc + c²) [∵ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)] = (b + c){a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²} – (b + c)(b² – bc + c²) = (b + c)[3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²] = (b + c)[3a² + 3ab + 3bc + 3ca] = 3(b + c)[a² + ab + bc + ca] = 3(b + c)[a(a + b) + c(a + b)] = 3(b + c)(a + b)(a + c) = 3(a + b)(b + c)(c + a) = R.H.S.

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

Answer»

We have,

L.H.S.

= (a + b + c)³ – a³ -b³ – c³

= {(a + b + c)³ – (a)³} – (b³ + c³)

= (a + b + c – a){(a + b + c)² + a(a + b + c) + a²} – (b + c){b² – bc + c²)

[∵ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)]

= (b + c){a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²} – (b + c)(b² – bc + c²)

= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²]

= (b + c)[3a² + 3ab + 3bc + 3ca]

= 3(b + c)[a² + ab + bc + ca]

= 3(b + c)[a(a + b) + c(a + b)]

= 3(b + c)(a + b)(a + c)

= 3(a + b)(b + c)(c + a)

= R.H.S.