Question

Solve the equation in the complex number system.

Answer 1

$x^3-1=0\\ (x-1)(x^2+x+1)=0\\ x-1=0\\ x=1\\ x^2+x+1=0\\ x^2+x+\frac{1}{4}+\frac{3}{4}=0\\ (x+\frac{1}{2})^2=-\frac{3}{4}\\ x+\frac{1}{2}=\sqrt{-\frac{3}{4}} \vee x+\frac{1}{2}=-\sqrt{-\frac{3}{4}}\\ x=-\frac{1}{2}+i\frac{\sqrt3}{2} \vee x=-\frac{1}{2}-i\frac{\sqrt3}{2}\\ x=-\frac{1-i\sqrt3}{2} \vee x=-\frac{1+i\sqrt3}{2}\\\\ x=\{1,-\frac{1-i\sqrt3}{2},-\frac{1+i\sqrt3}{2} \}$

Answer 2

To get rid of $x^{3}$, you have to take the third root of both sides:
$\sqrt[3]{x^{3}} = \sqrt[3]{1}$
But that won't help you with understanding the problem. It is better to write $x^{3}-1$ as a product of 2 polynomials:
$x^{3}-1 = (x-1)\cdot (x^{2} +x +1)$
From this we know, that $x-1 = 0 => x = 1$ is the solution. Another solutions (complex roots) are the roots of quadratic equation.