Explanation

(d)
Let $\alpha,\ \beta\ and\ \gamma$ be the roots of the given equation
$\therefore\ \alpha + \beta + \gamma = - 2,\ \alpha\beta + \beta\gamma + \gamma\alpha = - 4$
And $\alpha\beta\gamma = - 1$
Let the required cubic equation has the roots $3\alpha,\ 3\beta$ and $3\gamma$.
$\therefore 3\alpha + 3\beta + 3\gamma = - 6$
$3\alpha.3\beta + 3\beta.3\gamma + 3\gamma.3\alpha = - 36$
And $3\alpha.3\beta.3\gamma = - 27$
$\therefore$ Required equation is
$x^{3} - ( - 6)x^{2} + ( - 36)x - ( - 27) = 0$
$\Rightarrow x^{3} + 6x^{2} - 36x + 27 = 0$