<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] @user230452 asks for an enlightened way of doing this.
The expression 
$$
E = E(a, b, c) = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3
$$
is symmetric in $a, b, c$, and homogeneous of total degree $3$, so it must be an integral combination of the elementary symmetric polynomials in $a, b, c$, of the form
$$
E = x \sigma_{1}^{3} + y \sigma_{1} \sigma_{2} + z \sigma_{3},
$$
for suitable integers $x, y, z$. Here $\sigma_{1} = a + b + c$, $\sigma_{2} = ab + ac + bc$, $\sigma_{3} = a b c$ are the elementary symmetric polynomials in $a, b, c$.
Set in order 


*

*$a = b = c = 1$, 

*$a = b = 1$, $c = 0$, 

*$a = 1$, $b = c = 0$ 


to get
$$
\begin{cases}
3 x + 9 y + z &= E(1, 1, 1) &=24\\
2 x + 2 y &= E(1, 1, 0) &=0\\
x &= E(1, 0, 0) &=0\\
\end{cases}
$$
Hence $x = y = 0$, $z = 24$, and the expression evaluates to $$E = 24 \sigma_{3} = 24 a b c.$$
</s>