If $\alpha$ ,$\beta$are  roots of $4x^2+2x-1=0$, then $\left(4{\alpha }^3-3\alpha \right)$ 

Let the other root be $\beta$. Thus, $\alpha +\beta =-\frac{1}{2}$ and $\alpha \beta =-\frac{1}{4}$ Clearly, $4{\alpha }^2+2\alpha -1=0$

$\Rightarrow 4{\alpha }^2=1-2\alpha$ 

Now, 

$\begin{array}{l}4{\alpha }^3-3\alpha =\alpha \left(4{\alpha }^2-3\right)\\ =\alpha (1-2\alpha -3)\\ =-\alpha (2\alpha +2)\\ =-\left(2{\alpha }^2+2\alpha \right)\\ =-\frac{1}{2}\left(4{\alpha }^2+4\alpha \right)\\ =-\frac{1}{2}(1-2\alpha +4\alpha )\\ =-\frac{1}{2}(1+2\alpha )\\ =-\frac{1}{2}-\alpha =\beta \end{array}$

Hence, the result