Dodecahedron and golden ratio algebra

We can see that the volume of a dodecahedron of size $2\varphi$, where $\varphi=\frac{\sqrt{5}-1}{2}$, can be found in two ways. The first one uses the pentagonal pyramids with the faces as basis and height the distance of the face from the center of the dodecahedron and gives the result ( here $\Phi=\frac{1}{\varphi}=\varphi+1$): $$ V=4\sqrt{5(2-\varphi)}\sqrt{3-\frac{4\varphi^2}{4-\Phi^2}} $$ The other way start from a cube of side $a=2$ and add to every face a " roof" with penatagonal flaps, as in the figure, and gives the result $$V=8+4\Phi$$

Now I want to prove that $$ 4\sqrt{5(2-\varphi)}\sqrt{3-\frac{4\varphi^2}{4-\Phi^2}}=8+4\Phi $$ only with algebra, using the properties of Golden Ratio: $\varphi=1-\Phi=\frac{1}{\Phi}$

Too long for a comment, but this should be enough to help you answer the question yourself.

First, check your result numerically. Do you mean $8+4\Phi$?

Algebra with the golden ratio is really nice since any polynomial in $\Phi$ can be reduced to an equivalent degree 1 polynomial using the identity $\Phi^2 = \Phi+1$.

For instance: $$\Phi^3 = \Phi(\Phi^2) = \Phi(\Phi+1) = \Phi^2+\Phi = \Phi+1+\Phi = 2\Phi+1.$$ And: $$\Phi^4 = \Phi(\Phi^3)=\Phi(2\Phi+1) = 2\Phi^2+\Phi = 2(\Phi+1)+\Phi = 3\Phi+2.$$

So your identity can be proved by following these steps:

1. Eliminate $\phi$ from both sides of the equation: express everything in terms of $\Phi$
2. Square both sides to get rid of the square roots
3. Multiply both sides by $4-\Phi^2$ to clear the denominator
4. What remains should be some polynomial in $\Phi$ on the left and some polynomial in $\Phi$ on the right. Use $\Phi^2 = \Phi+1$ to reduce them to the form $a\Phi + b$. They should be equal.