If $\alpha$ is a root of 4x² + 2x- 1 = 0, then 4$\alpha$³ - 3$\alpha$ 

Since $\alpha$ is a root of $4x^2+2x-1=0$. Then $4{\alpha }^2+2\alpha -1=0$

$\begin{array}{l}\Rightarrow \alpha =\frac{-1\pm \sqrt{5}}{4}=\frac{\sqrt{5}-1}{4},\frac{-\sqrt{5}-1}{4}\\ \Rightarrow \alpha =\sin {18}^{\circ },-\sin {18}^{\circ }\end{array}$

Now, $4{\alpha }^3-3\alpha =4{\sin }^3{18}^{\circ }-3\sin {18}^{\circ }$

$\begin{array}{l}=-\sin {54}^{\circ }=-\cos {36}^{\circ }\\ =-\frac{\sqrt{5}+1}{4}\end{array}$