If one root of $4x^2+2x-1= 0$ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

If one root of $4x^2+2x-1= 0$ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

My solution:

By using the quadratic formula it says $\displaystyle x=\frac{-1+\sqrt{5}}{4}$ is a solution for $4x^2+2x-1= 0$ and it's also known to be $\cos 72^\circ$. The other solution for $4x^2+2x-1= 0$ is $\displaystyle x=-\left(\frac{1+\sqrt{5}}{4}\right)$ and it's known to be $-\cos 36^\circ=\cos (180^\circ+36^\circ)=\cos 3(72^\circ)$.

We therefore can conclude that if one of the roots for $4x^2+2x-1= 0$ is $\alpha$, which is $\cos 72^\circ$, the other root, $\cos 3(72^\circ)$ will be $4\alpha^3-3\alpha$, which abides by the triple angle formula for cosine function: $\cos 3\alpha=4\cos^3 \alpha -3\cos\alpha$.