## Algebra 2 (1st Edition)

$\dfrac{4}{3}, \dfrac{-2 +2\sqrt{3}i }{3}, \dfrac{-2 -2\sqrt{3} i}{3}$
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, $x=\dfrac{4}{3}, \dfrac{-2 +2\sqrt{3}i }{3}, \dfrac{-2 -2\sqrt{3} i}{3}$