<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] This technique comes from A Course in Pure Mathematics by Margaret M. Gow:
$E=(a+b+c)^{3}-(b+c-a)^{3}-(c+a-b)^{3}-(a+b-c)^{3}$
$c=0 \implies E=(a+b)^{3}-(b-a)^{3}-(a-b)^{3}-(a+b)^{3}=0$
$\therefore c$ is a factor of $E$.
By symmetry, $a, b$ also are factors of $E$.
Since $E$ is cubic, let $E=kabc$ where $k$ is a constant.
Take $a=b=c=1$, $E=24=k$.
$\therefore E=24abc$
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