Cubic

**franckherve1** · October 18th 2008, 06:23 PM

Let C be a cube where the length of its diagonal is the same as its volume..what's the length of each side?

**U-God** · October 18th 2008, 06:29 PM

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.

**franckherve1** · October 18th 2008, 06:34 PM

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

**mr fantastic** · October 18th 2008, 06:35 PM

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 ....

**U-God** · October 18th 2008, 06:37 PM

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: $\sqrt{ \sqrt{x^2 + x^2}^2 + x^2} = \sqrt{x^2 + x^2 + x^2}$

**franckherve1** · October 18th 2008, 06:43 PM

and how to solve (square root)3x(squared)=X(cubic)

**U-God** · October 18th 2008, 06:46 PM

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: $x = 3^{\frac{1}{4}}$

**franckherve1** · October 18th 2008, 06:49 PM

thanks so much!!

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