2016/035) If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

We have $4\alpha^2+2\alpha-1= 0\cdots(1) $
We need to prove
$4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
We have $4\alpha^2+2\alpha-1= 0$
Hence $4\alpha^3+2\alpha^2-\alpha= 0$
So $4\alpha^3 = -2\alpha^2 + \alpha$
or $4\alpha^3 -3\alpha = - 2\alpha^2 - 2 \alpha=  - \frac{1}{2}( 4\alpha^2 + 4\alpha)$
$= - \frac{1}{2}( 1 - 2\alpha + 4\alpha)= - \frac{1}{2}( 1 + 2\alpha) \dots(2)$
Hence $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
 = $4(\frac{1}{2}( 1 + 2\alpha))^2- 2 (\frac{1}{2}( 1 + 2\alpha)-1$ using (2)
 = $( 1 + 2\alpha))^2- ( 1 + 2\alpha)-1$
 = $1 + 4 \alpha + 4\alpha ^2 - 1 -2 \alpha -1$
 = $4\alpha ^2 + 2 \alpha -1= 0$ from (1)