Volume of a cube inscribed in a cone. Solve this ratio?

Question

A right circular cone has a cube inscribed in it. (This means that the four upper vertices of the cube touch the inside of the cone at an unknown point) If the radius of the base of the cone is 1 and its height is 3, what is the volume of the cube?

Now, I'm using the similar triangles method. The smaller triangle has a height of 3-x, and the width is x(root)2. The larger triangle has a length of 3 and a width of 2. Can anyone solve the following ratio? (I'm horrible with algebra)

(3-x) / x(root)2 = 3 / 2

Answer 1

(3-x) / x(root)2 = 3 / 2

2(3 - x) = 3(√2 x)

6 - 2x = 3√2 x

x(2 + 3√2) = 6

x = 6 / (2 + 3√2)

= 0.961131723

volume of inscribed cube

= x^3

= 0.887868678 cu