Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a).
Solution: L.H.S. = (a + b + c)3 – a3 – b3 – c3 = {(a + b + c)3 – 3} – {b3 + c3} = (a + b + c – a) {(a + b + c)2 + a2 + a(a + b + c)} – (b + c) (b2 + c2 – bc) = (b + c) {a2 + b2 + 2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac – b2 – a2 + bc) = (b + c) (3a2 + 3ab + 3bc + 3ca} = 3(b + c) {a2 + ab + bc + ca} = 31b + c) {{a2 + ca) + (ab + bc)} = 3(b + c) {a(a + c) + b(a + c)} = 3(b + c)(a + c) (a + b) = 3(a + b)(b + c) (c + a) = R.H.S.
