Answer
$a^{3}x^{3}+b^{3}=0$
$(ax)^{3}+b^{3}=0$
$(ax+b)\times((ax)^{2}-axb+b^{2})=0$
$x=\frac{-b}{a}$.
$x=\frac{b(1+i\sqrt 3)}{2a}$.
$x=\frac{b(1-i\sqrt 3)}{2a}$.
Work Step by Step
$a^{3}x^{3}+b^{3}=0$
$(ax)^{3}+b^{3}=0$
$(ax+b)\times((ax)^{2}-axb+b^{2})=0$
$ax+b=0$ or $(ax)^{2}-axb+b^{2})=0$
With $ax+b=0$, then $x=\frac{-b}{a}$
With $((ax)^{2}-abx+b^{2})=0$, then $x=\frac{ab+\sqrt ((-ab)^{2}-4a^{2}b^{2})}{2(a)^{2}}$ and $x=\frac{ab-\sqrt ((-ab)^{2}-4a^{2}b^{2})}{2(a)^{2}}$.
With $x=\frac{ab+\sqrt ((-ab)^{2}-4a^{2}b^{2})}{2(a)^{2}}=\frac{ab+\sqrt (-3a^{2}b^{2})}{2(a)^{2}}=\frac{ab+\sqrt (3a^{2}b^{2}i^{2})}{2(a)^{2}}=\frac{ab+abi\sqrt 3}{2(a)^{2}}=\frac{b(1+i\sqrt 3)}{2a}$.
With $x=\frac{ab-\sqrt ((-ab)^{2}-4a^{2}b^{2})}{2(a)^{2}}=\frac{ab-\sqrt (-3a^{2}b^{2})}{2(a)^{2}}=\frac{ab-\sqrt (3a^{2}b^{2}i^{2})}{2(a)^{2}}=\frac{ab-abi\sqrt 3}{2(a)^{2}}=\frac{b(1-i\sqrt 3)}{2a}$.
