Question

Prove that (a+b)cube+(b+c)cube+(c+a)cube-3(a+b)×(b+c)×(c+a)=2(acube+bcube+ccube)

Answer 1

Taking RHS of the identity:
(a + b + c)(a2 + b2 + c2 - ab - bc - ca )

Multiply each term of first polynomial with every term of second polynomial, as shown below:
= a(a2 + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 - ab - bc - ca ) + c(a2 + b2 + c2 - ab - bc - ca )

= { (a X a2) + (a X b2) + (a X c2) - (a X ab) - (a X bc) - (a X ca) } + {(b X a2) + (b X b2) + (b X c2) - (b X ab) - (b X bc) - (b X ca)} + {(c X a2) + (c X b2) + (c X c2) - (c X ab) - (c X bc) - (c X ca)}

Solve multiplication in curly braces and we get:
= a3 + ab2 + ac2 - a2b - abc - a2c + a2b + b3 + bc2 - ab2 - b2c - abc + a2c + b2c + c3 - abc - bc2 - ac2

Rearrange the terms and we get:
= a3 + b3 + c3 + a2b - a2b + ac2- ac2 + ab2 - ab2 + bc2 - bc2 + a2c - a2c + b2c - b2c - abc - abc - abc

Above highlighted like terms will be subtracted and we get:
= a3 + b3 + c3 - abc - abc - abc

Join like terms i.e (-abc) and we get:
= a3 + b3 + c3 - 3abc

Hence, in this way we obtain the identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Answer 2

I think it should be answer.