Question

prove that (a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)

Answer 1

LHS=
$(a + b + c) {}^{3} - a {}^{3} - b {}^{3} - c {}^{3}$
We have to expand a+b+c whole cube. So,
$= a {}^{3} + b {}^{3} + c {}^{3} + 3(a + b)(b + c) \\ (c + a) - a {}^{3} - b {}^{3} - c {}^{3}$
Cancelling out (a) cube (b) cube and (c) cube.
$= 3(a + b)(b + c)(c + a)$
LHS=RHS
--------PROVED----------