Answer
$\dfrac{4}{3}, \dfrac{-2 +2\sqrt{3}i }{3}, \dfrac{-2 -2\sqrt{3} i}{3}$
Work Step by Step
Re-write as: $27x^3-(4)^3=0 \implies (3x-4)(9x^2+12x+16)=0$
and $3x-4 =0 \implies x=\dfrac{4}{3}$
Solving the equation $9x^2+12x+16$ for $x$, we find:
$x=\dfrac{-(-12)+\sqrt{(12)^2-(4)(9)(16)}}{(2)(9)}= \dfrac{-2\pm 2\sqrt{3}}i{3}$
So, $x=\dfrac{4}{3}, \dfrac{-2 +2\sqrt{3}i }{3}, \dfrac{-2 -2\sqrt{3} i}{3}$
