Question

(a+b)^3+(b+c)^3+(c+a)^3-3(a-b)(b-c)(c-a)=2(a^3+b^3+c^3-3abc)

Answer 1

I believe there is a typo, it should have read:

Show that
(a+b)^3+(b+c)^3+(c+a)^3-3(a + b)(b + c)(c + a) = 2(a^3+b^3+c^3-3abc)

For the first three terms, use standard identities:
(a+b)³=a²+3a²b+3ab²+b³
(b+c)³=b²+3b²c+3bc²+c³
(c+a)³=c²+3c²a+3ca²+a³

For the fourth term, do FOIL and multiply again:
Y=(a+b)(b+c)(c+a)
=(ab+ac+b²+bc)(c+a)
=abc+a²b+ac²+a²b+b²c+b²a+bc²+abc

If we multiply Y by -3 and add to the sum of the first three cubes, the cross terms all cancel, leaving behind the right hand side of the identity.