View Full Version : Any ideas on how to get further in this intro. math proof?

josephdinatale

02-13-2011, 10:49 AM

I can't seem to get any further right now in this proof:

Result: Let a, b, and c be the lengths of the sides of a triangle T, where a ≤ b ≤ c. If T is a right triangle, then (abc)^2 = (c^6 – a^6 – b^6) / 3.

Proof: Assume T is a right triangle. Assume the pythagorean theorem. Then a and b are the lengths of the legs of the right triangle and c is the length of the hypotenuse. Thus a^2 + b^2 = c^2.

Then c = sqrt(a^2 + b^2)

Then (abc)^2 = (a^4)(b^2) + (b^4)(a^2)

But I have no idea how to get further so that there is an exponent to the power 6 and so that there is division by 3.

Any tips?

Modiga-Disabled

02-13-2011, 10:56 AM

I believe it has something to do with binomial expansion.

Consider (s+t)^3 = s^3 + 3(s^2)*t + 3s(t^2) + t^3

Look for similarities between that and (abc)^2 = (a^4)(b^2) + (b^4)(a^2)

josephdinatale

02-13-2011, 11:48 AM

I believe it has something to do with binomial expansion.

Consider (s+t)^3 = s^3 + 3(s^2)*t + 3s(t^2) + t^3

Look for similarities between that and (abc)^2 = (a^4)(b^2) + (b^4)(a^2)

This is just one way I factored (abc)^2, it was a wild guess, and I'm not sure if its on the right path. Do you think that its going in the right direction?

(abc)^2 = (a^4)(b^2) + (b^4)(a^2)

Modiga-Disabled

02-13-2011, 12:46 PM

This is just one way I factored (abc)^2, it was a wild guess, and I'm not sure if its on the right path. Do you think that its going in the right direction?

It's the right direction.

The next step is to go from there using (s+t)^3 = s^3 + 3(s^2)*t + 3s(t^2) + t^3.

To give a further hint, substitute in s=a² and b=² and do some rearranging. You should be able to get one side of that equation equal to what you already have.

josephdinatale

02-13-2011, 01:46 PM

It's the right direction.

The next step is to go from there using (s+t)^3 = s^3 + 3(s^2)*t + 3s(t^2) + t^3.

To give a further hint, substitute in s=a² and b=² and do some rearranging. You should be able to get one side of that equation equal to what you already have.

Thank you, I completed the proof

Do you think it would be academic dishonesty if I submitted my proof after using your hints? Would I need to reference you in any way?

cdemonkey

02-13-2011, 01:47 PM

Try asking the WoW support staff.

josephdinatale

02-13-2011, 02:15 PM

Try asking the WoW support staff.

haha i get the reference.

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