If $\alpha$ be a root of the equation $4x^2+2x-1=0$, prove that $4{\alpha }^3-3\alpha$ is the other root.

Given $4x^2+2x-1=0$ roots are $\alpha \text{\hspace{0.17em}and\hspace{0.17em}}\beta$

$\alpha +\beta =\frac{-b}{a}=\frac{-2}{4}$

$\alpha +\beta =\frac{-1}{2},\alpha \beta =\frac{c}{a}=\frac{-1}{4}$

Also $4{\alpha }^2+2\alpha -1=0$ as $\alpha$ is a root and we have to prove that

$\beta =4{\alpha }^3-3\alpha$

$=4{\alpha }^2\cdot \alpha -3\alpha$

$=\alpha (1-2\alpha )-3\alpha$

$=-2{\alpha }^2-2\alpha =\frac{-1}{2}[4{\alpha }^2+4\alpha ]$

$=\frac{-1}{2}[1-2\alpha +4\alpha ]$

$=\frac{-1}{2}\left[(1+2\alpha )\right]=\frac{-1}{2}-\alpha =\beta$

$\alpha +\beta =\frac{-1}{2}$ by (1)

Hence, the other root $\beta$ is $4{\alpha }^3-3\alpha$.