How to solve $x^3 -1 =0$

Question

I know what the roots are - according to this page: How to solve $x^3 = 1$?

But i don't understanding why the roots are: $e^{\frac23i\pi}$ and $e^{\frac43i\pi}$.

Answer 1

Whenever you want to solve an equation, you need to be precise about the field in which you are looking for the roots. In general, unless you are in $\mathbb{C}$ where any polynomial has as many roots as its degree, an equation may have no root (example $x^2-2$ over $\Bbb{Q}$) one root (example $x^3-1$ over $\Bbb{R}$)...

In your case when looking to complex roots of the form $x=\rho\cdot e^{i\theta}$ one has $x^3=1=\rho^3\cdot e^{3i\theta}$ and this means besides $\rho=1$

$$3i\theta=2k\pi\,\,\,(k=0,1,2)$$

And this gives three distinct solutions $1,e^{2i\pi/3},e^{4i\pi/3}$

Answer 2

Because of $e^{i2\pi}=1$ and therefore also $e^{i4\pi}=1$ the roots are solving the equation. Think about $(-1)\cdot(-1)=1\cdot 1$, that's the same type of problem.