Factorization of three variables Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) $
Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - [(c+a)^3 +(a+b)^3 +(b+c)^3]$
Putting $A:=(c+a)$ ; $B:=(a+b)$ ; $C:=(b+c)$
$[(c+a)^3 +(a+b)^3 +(b+c)^3] $ becomes $(A^3+B^3+C^3)$ now again using the formulae:$ a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(b+c)(c+a)$ 
Could you please guide if any other easier way of doing this...
 A: If you simply expand everything and collect terms, everything will cancel out and reduce down to 6abc. However, if you want to factorize it, think about the other terms you have - 
As you said, $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. So then what does $(a+b)^3$ equal? If you expand it out following the same format as $(a+b+c)^3$ you get $a^3+b^3+3ab(a+b)$
So now your expression becomes 
$$a^3+b^3+c^3+3(a+b)(b+c)(c+a)-a^3-b^3-3ab(a+b)-b^3-c^3-3bc(b+c)-a^3-c^3-3ac(a+c)+a^3+b^3+c^3$$
Clean up all the cubes in there and you get
$$3(a+b)(b+c)(c+a)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$
Expand the first part of that to get
$$3(2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$
Which becomes
$$3(2abc+ab(a+b)+bc(b+c)+ac(a+c))-3(ab(a+b)+bc(b+c)+ac(a+c))$$
Now just cancel out all those like terms and you get $6abc$
A: OK - if you need a really shorter way (which may work depending on the equation you have):  
You have $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$  
Looking at the LHS as a cubic polynomial in $a$, we immediately can notice $a=0$ is a root and hence $a$ is a factor.  By symmetry, $b$ and $c$ are factors. As $abc$ must be a factor of the LHS (which is homogeneous and also of degree $3$), the only other factor possible is a scalar. Hence LHS is of form $k\cdot abc$ and the scalar $k$ can be obtained by checking for easy values, say $a = b = 1$ and $c = -1$.
