# Cubic

• Oct 18th 2008, 07:23 PM
franckherve1
Cubic
Let C be a cube where the length of its diagonal is the same as its volume..what's the length of each side?
• Oct 18th 2008, 07:29 PM
U-God
Well let the length of one side = x.

Volume = $x^3$

length of diagonal = $\sqrt{x^2 + x^2 + x^2}$ from Pythagoras' theorem.

Let them equal each other.
• Oct 18th 2008, 07:34 PM
franckherve1
shouldn't be like square root (x2+x2) ...why for the diagonal..its' three times?
• Oct 18th 2008, 07:35 PM
mr fantastic
Quote:

Originally Posted by franckherve1
shouldn't be like square root (x2+x2) ...why for the diagonal..its' three times?

The diagonal length of the base of the cube is $\sqrt{x^2 + x^2}$. Therefore ....
• Oct 18th 2008, 07:37 PM
U-God
Well it's a cube isn't it? It's a three dimensional object. $\sqrt{x^2 + x^2}$ will give you the hypotenuse of two dimensions, but that is the diagonal along one face of the cube, from your question, I assumed you wanted the diagonal between opposing points of the cube, ie. front bottom left, to top right back.

in which case you will use the first hypotenuse and use Pythagoras' theorem with another edge piece to find the next hypotenuse (the answer you're searching for)

This is: $\sqrt{ \sqrt{x^2 + x^2}^2 + x^2} = \sqrt{x^2 + x^2 + x^2}$
• Oct 18th 2008, 07:43 PM
franckherve1
and how to solve (square root)3x(squared)=X(cubic)
• Oct 18th 2008, 07:46 PM
U-God
Well, $\sqrt{3x^2} = x^3$

square both sides, $3x^2 = x^6$

Then divide both sides by x squared $3 = x^4$

Then take a quartic root of both sides: $x = 3^{\frac{1}{4}}$
• Oct 18th 2008, 07:49 PM
franckherve1
thanks so much!!