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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...
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 $6abc$
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 $c = -1$.
