<s>[INST] Find four integers summing to zero, with sum of cubes 24 I'm stuck on the following problem from Terence Tao's "Solving Mathematical Problems"

Find all integers $a,b,c,d$ such that $a+b+c+d=0$ and $a^3+b^3+c^3+d^3=24$. (Hint: it is not hard to guess some solutions ti these equations, but to show that you have all of them, substitute the first equation into the second and factorize.)


My attempt:
After said substitution, I now want to solve
$$-3 a^2 b-3 a^2 c-3 a b^2-6 a b c-3 a c^2-3 b^2 c-3 b c^2-24=0 $$
by factoring the LHS. There is some evident symmetry, so I tried going with $c=Aa+Bb+C$ for one of the factors without success. Moreover, Mathematica can't seem to factorize this polynomial as well.
I'd appreciate any help on this problem, thanks!
 [/INST] Divide by $-3$ of your expression, we have
$$a^2 b+ a^2 c+ a b^2+2 a b c+ a c^2+ b^2 c+ b c^2+8=0 $$
Note that 
$$\begin{split}(a+b)(b+c)(c+a) &= (ab+ac + b^2 +bc)(c+a)\\
&= abc + ac^2 + b^2c + bc^2 + a^2b + a^2 c+ ab^2 + abc \\
&= -8
\end{split}$$ 
As $a, b, c$ are integers, there are not many choices. You need then to solve 
$$\begin{split} a + b &= \alpha \\ b+c &= \beta \\ a+c &= \gamma \end{split}$$
$$\Rightarrow a = \frac{\alpha -\beta +\gamma}{2},\ \ b = \frac{\alpha + \beta - \gamma}{2},\ \ c = \frac{-\alpha + \beta + \gamma}{2}. $$
where $\alpha, \beta, \gamma$ are any integers so that $\alpha\beta\gamma = -8$. Now you need only to check which choices of $\alpha, \beta, \gamma$ would make $a, b, c$ integers. 
(To come up with that factorization, one observe that the expression is of homogeneuos degree three (except the constant term) and are highly symmetric in $a, b, c)$. Thus $(a+b)(b+c)(c+a)$ is more or less the first thing you want to try.)
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