Answer
$\bigg\{ -\sqrt[3]2,-1, \frac{\sqrt[3]2\pm i\sqrt{3\sqrt[3]4}}{2}, \frac{1\pm i\sqrt 3}{2}\bigg\}$
Work Step by Step
$x^6+3x^3+2=0$
Let $x^3=a$
$a^2+3a+2=0,$
$a^2+a+2a+2=0,$
$a(a+1)+2(a+1)=0,$
$(a+2)(a+1)=0,$
Substituting $x^3=a$
$(x^3+2)(x^3+1)=0$
$(x+\sqrt[3]2)(x^2-\sqrt[3]2x+\sqrt[3]4)(x+1)(x^2-x+1)=0$
The solutions are:
$x_1=-\sqrt[3]2$
$x_2=-1$
$x_{3,4}=\frac{\sqrt[3]2\pm i\sqrt{3\sqrt[3]4}}{2}$
$x_{5,6}=\frac{1\pm i\sqrt 3}{2}$
Thus the solution set is:
$\bigg\{ -\sqrt[3]2,-1, \frac{\sqrt[3]2\pm i\sqrt{3\sqrt[3]4}}{2}, \frac{1\pm i\sqrt 3}{2}\bigg\}$
