Answer: (4)

Solution

Let $\alpha$ and $\beta$ be the roots of the equation, $81x^2 + kx + 256 = 0$

Given $(\alpha)^3 = \beta$ (wait, actually $\alpha = \beta^3$ or $\beta = \alpha^3$)

$\alpha \cdot \beta^3 = \frac{256}{81}$ (product) and $\alpha = \beta^3$

$\therefore (\alpha)(\beta) = \frac{256}{81}$

$\Rightarrow \beta^4 = \left(\frac{4}{3}\right)^4 \Rightarrow \beta = \frac{4}{3}$

$\Rightarrow \alpha = \frac{64}{27}$

$\therefore$ Sum of the roots $= \frac{-k}{81}$

$\therefore \alpha + \beta = \frac{-k}{81} \Rightarrow \frac{4}{3} + \frac{64}{27} = \frac{-k}{81}$

$\Rightarrow k = -300$