# A^3 + b^3 = c^3 where a, b, c, are sides of a triangle?

1. Apr 5, 2009

### InigoMontoya

Hi,

I'm doing a project for math class on Fermat's Last Theorem, and the first thing I wanted to know when I learned what the theorem was, was how a3+b3 = c3 would look in relation to the way a2+b2 = c2 is often depicted. That is, I wanted to know if there was a way to show Fermat's equation to the 3rd power (with a, b, and c as real numbers, not integers, obviously) in a way similar to this:

http://www.clascalc.com/Images/pythagorean-theorem-1.gif [Broken]

What I came up with last night is depicted in the attached image at the bottom of this post (sorry, I did it quickly in MS paint.. but you get the idea, I think).

What I want to know now is, is my diagram incorrect? Am I wrong in assuming that the cube root of a3+b3 is equal to c when c is the hypotenuse of a right-angled triangle? In other words, sure, you can build three cubes off the sides of a two-dimensional triangle, and you can also measure the sides of that triangle and plug two of the measurements into the equation a3+b3 = c3, but the result given in the equation will not be equal to the measurement you took of the third side?

Does anyone know whether this is true or false? And why? I realize I probably haven't done an excellent job of explaining myself, but I tried my best. I can't seem to find any answers on Google, and I'm having trouble figuring it out for myself - I think I'm just a tiny bit (no sarcasm) out of my depth in this topic (it's meant to be an enrichment project, after all). I did build three cubes out of bristol board today based on an equation I did where 83+83 = c3 (c = 10.079...), but when I tried to put them together as a right-angled triangle, the 'c' cube, or hypotenuse, was too short. It could have just been an error in precision on my part, or it could be an error in the mathematical reasoning. I would be extremely grateful if someone could let me know which one it was.

Thanks!

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• ###### 00cubes.GIF
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2. Apr 5, 2009

### mathman

a^2 + b^2 = c^2 is the Pythagorean theorem where c is the hypotenuse of a right triangle. a^3 + b^3 = c^3 will hold for some triangle, but it won't be a right triangle.
Fermat's last theorem has nothing to do with it.

3. Apr 6, 2009

### InigoMontoya

Hmmm... I see. Thank you.

I don't know if I'm allowed to ask this here, but my question that arises from that would be, why does a2+b2 = c2 always form a right triangle? Why not some other triangle?

4. Apr 6, 2009

### m00npirate

Think trig identities. sin^2x + cos^2x = 1. what do sinx and cosx mean?

5. Apr 7, 2009

### Redbelly98

Staff Emeritus
One way to answer your question is to say, "It's true because _____", where "_____" is any one of the 81 proofs of the theorem given here:

http://www.cut-the-knot.org/pythagoras/index.shtml
(Scroll down 3 or 4 screens to see the 1st proof.)

While I'm no expert, my recollection is that the Pythagorean theorem is used to prove that relation. So to use it to "prove" Pythagoras would be circular reasoning.

6. Apr 7, 2009

### matt grime

There is more than one way to prove almost anything, and there is more than one way to show that the trig identity given is true, so it is not really circular reasoning to invoke it.

7. Apr 7, 2009

### Gokul43201

Staff Emeritus
Strictly speaking, Inigo is asking for a proof of the converse of the Pythagoras Theorem. But that's just one step removed (if you accept - without proof - a congruence theorem for triangles) from the proof of the theorem itself.

Last edited: Apr 7, 2009
8. Apr 15, 2009

### MLeszega

Isn't it because this is just a special case of the law of cosines? If you didn't have a right triangle you would simple have c^2 = a^2 + b^2 - 2abcos(theta). Cos(90) = 0, so you have c^2 = a^2 + b^2 - 0 => c^2 = a^2 + b^2