<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] with $$c^2=a^2+b^2$$ the left-hand side is given by
$$(ab\sqrt{a^2+b^2})^2$$ and the other side
$$\frac{(\sqrt{a^2+b^2})^6-a^6-b^6}{3}$$
can you compute this?
expanding the right-hand side we get $$a^4b^2+a^2b^4$$ and so is the left-hand side.
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