<s>[INST] Proof that $(abc)^2 = \frac{(c^6-a^6-b^6)}{3}$ We are given the length of the sides of a right triangle T, where $a \leq b \leq c$. We are asked to prove if $T$ is a right triangle. then $(abc)^2 = (c^6-a^6-b^6)/3$.
What I tried:
I tried substituting with the Pythagorean theorem $a^2 + b^2 = c^2$.
Where I am stuck:
I am unable to get $\frac{1}{3}$ or $3$ anywhere in my simplification. Is there some kind of identity that I am missing?
 [/INST] Hint : Cube the equation $\color{blue}{a^2+b^2=c^2}$
\begin{eqnarray*}
a^6+b^6+\color{red}{3} a^2 b^2 (\color{blue}{a^2+b^2}) =c^6.
\end{eqnarray*}
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