If one real root of the quadratic equation $81x^2+kx+256=0$ is cube of the other root, then a value of $k$ is :

The correct option is A $-300$
Let one root of the given quadratic equation be $\alpha$.
Then the other root is ${\alpha }^3$
$\Rightarrow {\alpha }^4=\frac{256}{81}$
$\Rightarrow \alpha =\frac{4}{3}$
$\frac{-k}{81}=\alpha +{\alpha }^3=\left(\frac{4}{3}\right)+{\left(\frac{4}{3}\right)}^3$
$\Rightarrow k=-300$