Prove that if T is a right triangle then (abc)^2 = c^6 - a^6 - b^6/3?

If T is a right triangle, and a ≤ b ≤ c, then

Case 1: if a is a hypotenuse, then
a ≠ b and a ≠ c as there is one, and only one hypotenuse in a right triangle, and
a is the greatest, hence
a > b and a > c, but this contradicts the assumption, so
a is not the hypotenuse.

Case 2: if b is a hypotenuse, then
b ≠ a and b ≠ c as there is one, and only one hypotenuse in a right triangle, and
b is the greatest, hence
b > a and b > c, but "b > c" contradicts the assumption, so
b is not the hypotenuse.

From case 1 and 2, c is the hypotenuse, and
a ≤ b < c.

Thus,
a^2 + b^2 = c^2 (Pythagorean Theorem)
(a^2 + b^2)^3 = (c^2)^3
a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = c^6
3a^4b^2 + 3a^2b^4 = c^6 - a^6 - b^6
a^4b^2 + a^2b^4 = (c^6 - a^6 - b^6)/3
a^2b^2(a^2 + b^2) = (c^6 - a^6 - b^6)/3
(abc)^2 = (c^6 - a^6 - b^6)/3

I hope this helps
now there is a formula that a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc
(abc)^2 = (c^6 - a^6 - b^6)/3

3(abc)^2 = (c^2)^3+(-b^2)^3+(-a^2)^3

applying the above formula

3(abc)^2=(c^2 - b^2 - a^2) (c^4 + b^4 + a^4 – a^2b^2 + b^2c^2 – c^2a^2) + 3(abc)^2

]3(abc)^2 terms get cancels from both sides

(c^2 - b^2 - a^2) (c^4 + b^4 + a^4 – a^2b^2 + b^2c^2 – c^2a^2)=0

either (c^4 + b^4 + a^4 – a^2b^2 + b^2c^2 – c^2a^2)=0 or (c^2 - b^2 - a^2)=0
taking the second condition
(c^2 - b^2 - a^2)=0
c^2 = b^2 + a^2 which is the pythagoras theorem which is applicable for right triangles only
hence T is a right triangle