Answer
$\bigg\{-2,1,-1\pm i\sqrt 3,\frac{-1\pm i\sqrt 3}{2}\bigg\}$
Work Step by Step
$x^6+7x^3-8=0$,
Let $x^3=a$.
$a^2+7a-8=0,$
$a^2-a+8a-8=0,$
$a(a-1)+8(a-1)=0,$
$(a+8)(a-1)=0,$
Substituting $x^3=a$
$(x^3+8)(x^3-1)=0$
$(x+2)(x^2-2x+4)(x-1)(x^2+x+1)=0$
The solutions are:
$x+2=0\Rightarrow x_1=-2$
$x-1=0\Rightarrow x_2=1$
$x_{3,4}=\frac{-1\pm i\sqrt 3}{2}$
$x_{5,6}=1\pm i\sqrt 3$
The solution set is $\bigg\{-2,1,-1\pm i\sqrt 3,\frac{-1\pm i\sqrt 3}{2}\bigg\}$.
Check the solutions:
$(-2)^6+7(-2)^3-8=0$
$1^6+7(1^3)-8=0$
$\left(\frac{-1\pm i\sqrt 3}{2}\right)^6+7\left(\frac{-1\pm i\sqrt 3}{2}\right)^3-8=0$
$(-1\pm i\sqrt 3)^6+7(-1\pm i\sqrt 3)^3-8=0$
