Squares difference simplifications

Question

I need to simplify this expression

$(a+b+c)^4-(a+b)^4-(a+c)^4-(b+c)^4+a^4+b^4+c^4$

Calculating each one will lead to over 30 different items.I observe that the final form should have a,b,c,d at the power of 4 but I don't know how the other items will look like.

I tried as $((a+b+c)^2)^2-((a+b)^2)^2-(a+c)^4-(b+c)^4+a^4+b^4+c^4$

but it doesn't help a lot.

Answer 1

The closed form is $12{abc}{(a+b+c)}$. Here is what I think, plug a=0 into the equation and we get the answer is 0. This implies that a is a factor of the final term. Since a, b, c are symmetric, the final term has abc in it. And if we plug a+b+c=0 into the equation, we also get a zero. Therefore the term (a+b+c) is also in it. Hence, the final term is $abc*(a+b+c)$ times a constant. And we can find out that the final term has ${3}^{4}-{3}*{2^4}+3=36$ terms, therefore we can derive that the constant must be 12.

Answer 2

The expression cancels if $a$, $b$ or $c$ is $0$. It also cancels if $a+b+c=0$.

So, as an homogeneous symmetric polynomial of degree $4$, it has to be a multiple of $abc(a+b+c)$. All you have to do is find the constant before $a^2bc$.

All this, of course, because I watched the answer in a CAS :-)