<s>[INST] 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...
 [/INST] 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 $c = -1$.
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