<s>[INST] Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even 
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even

I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this without checking every combination of the parity of $a,b,c$?
 [/INST] We will use that for any $A$ the numbers $A$, $3A$ and $A^3$ all have the same parity. Then writing $S$ for our expression, we see
$$\begin{align*}
S &= ab(a+b)+bc(b+c)+ca(c+a) \\ &\equiv 3ab(a+b) + 3bc(b+c)+3ca(c+a) \\
 &= (a+b)^3 - a^3 - b^3 + (b+c)^3 - b^3 - c^3 + (c+a)^3 - c^3 - a^3 \\
 &\equiv (a+b)^3 + (b+c)^3 + (c+a)^3 \\
 &\equiv (a+b) + (b+c) + (c+a) \equiv 0.
\end{align*}
$$
Here $\equiv$ denotes equality modulo $2$, i.e. $M \equiv N$ means that $M$ and $N$ have the same parity. So $S$ has the same parity as $0$, hence it is always even.
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