My proof involving Pythagorean’s Theorem

[Jadehaan]

Homework Statement

Let a, b, and c be lengths of sides of triangle T, where a ≤ b ≤ c.
Prove that if T is a right triangle, then (abc)²=(c⁶-a⁶-b⁶)/3

Homework Equations

If T is a right triangle, then Pythagorean’s Theorem states:
The sum of the squares of the lengths of the sides of a right triangle is equal to the square of the length of the hypotenuse. That is a²+b²=c², where c is the hypotenuse.

The Attempt at a Solution

We assume the given equation and using Pythagorean’s Theorem, we obtain solutions for c² and c⁶:
We substitute these results into the original equation.
This produces an equation where the left hand side is identical to the right hand side.
Since these terms are equal, it follows that the original equation holds true for a right triangle.

This is what I have. I am curious to if the proof is correct/acceptable.

Thanks for any feedback.

 

[snipez90]

Can you be more specific? It sounds like you're plugging in what you find into the identity you are trying to prove, which is flawed in that it assumes your identity is true in the first place. You want to start with the expression on one side of the identity and get to the other expression. (You keep referring to the "original equation", which I take to be the identity you are trying to prove, and not the pythagorean identity).

For instance, take the pythagorean identity and raise both sides to the third power. Subtract a^6 and b^6 from both sides to get an expression for c^6 - a^6 -b^6. Divide by 3, factor, and use the pythagorean identity again to get the left hand side of the identity you are trying to prove.

 

[Dick]

You didn't really show what you did. But if you substituted c=sqrt(a^2+b^2) and got the same result on both sides, I think that's just fine.

 

[Jadehaan]

Thanks for the help snipez90 and Dick. I redid the problem.

This is my original:

And this is my new proof:

 

[jgens]

Your first proof wasn't really a proof at all since you assumed what you wanted to prove and then arrived at a familiar equality (although, this can be a useful strategy if every step that you take is reversible). Your second proof looks correct to me however.