Explanation

(a)
Let $x = ( - 1)^{1/3}$
$x = \left( \cos{\pi + i\sin\pi} \right)^{1/3}$
$x = \left\lbrack \cos\left( \frac{2n + 1}{3} \right)\pi + i\sin{\left( \frac{2n + 1}{3} \right)\pi} \right\rbrack = e^{i(2n + 1)\pi/3}$
Put $n = 0,\ 1,\ 2$ we get
$x = e^{i\pi/3},\ e^{i\pi},\ e^{5i\pi/3}\$
$\therefore$ Products of roots$= e^{i\pi/3},\ e^{\pi i}.e^{5\pi i/3} = e^{3\pi i}$
$= \left( \cos{3\pi + i\sin{3\pi}} \right) = - 1$
Alternate Method
We know that the cube roots of -1 are -1, -$\omega,\ - \omega^{2}$
$\therefore$ Their product $= ( - 1)( - \omega)\left( - \omega^{2} \right) = - 1$