Explanation

(d)
Put $x^{1/3 =} = y$, then
$y^{2} + y - 2 = 0$
$\Rightarrow \ \ y = 1$ or $y = - 2$
$\Rightarrow x^{1/3} = 1$ or $x^{1/3} = - 2$
$\therefore\ \ \ \ x = (1)^{3}$ or $x = ( - 2)^{3} = - 8$
Hence, the real roots of the given equation are $1,\ - 8$