<s>[INST] Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of 

$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$

the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but can not proceed further :(
 [/INST] Let $\displaystyle b+c-a=x,c+a-b=y,a+b-c=z\implies x+y+z=a+b+c$
We need $\displaystyle(x+y+z)^3-x^3-y^3-z^3=\{x+(y+z)\}^3-x^3-y^3-z^3$
$\displaystyle=x^3+(y+z)^3+3x(y+z)\{x+(y+z)\}-x^3-y^3-z^3$
$\displaystyle=3yz(y+z)+3x(y+z)\{x+(y+z)\}$
$\displaystyle=3(y+z)\{yz+x(x+y+z)\}$
$$\displaystyle\implies(x+y+z)^3-x^3-y^3-z^3=3(y+z)(z+x)(x+y)$$
Now $y+z=2a$ etc.
Can you take it from here?
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