Complex solutions of a third degree equation

Question

I have to find the solutions of this third degree equation $x^3-2=0$. Obviously one of them is $\sqrt[3]{2}$, but i don't know how to to find the other two complex solutions. Any hints?

Answer 1

Since $\sqrt[3] 2$ is a root of $f(x)=x^3-2$, the polynomial $x-\sqrt[3] 2$ divides $f(x)$. So $f(x)=(x-\sqrt[3] 2)(x^2+ax+b)$ for some real numbers $a,b$. If you perform the long division you can obtain $a$ and $b$, and then solve the quadratic equation $x^2+ax+b=0$ using the quadratic formula, which gives the two complex roots.

Here is a more general solution:

If $f(x)=x^n-a$ then the roots of $f(x)$ are $\sqrt[n] a,~ \alpha\sqrt[n] a, \ldots, ~\alpha^{n-1}\sqrt[n] a$ where $\alpha$ is a primitive $n$-th root of unity (e.g. $e^{2\pi i/n}$).

Answer 2

Hint: Write

$$x^3-(\sqrt[3]{2})^3$$ and use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

We go to polar coordinates, keep track of the cyclic nature of the complex exponential function and pull the root there: \begin{align} z^3 &= 2 \iff \\ (r e^{i\phi})^3 &= 2 \, e^{i\cdot (0 + 2\pi k)} \iff \\ r e^{i \phi} &= \sqrt[3]{2} \, e^{i \cdot 2\pi k / 3} \end{align} Here $k \in \mathbb{Z}$.

We compare and get radius and three different angles (up to $2\pi$ cycles): $$ r = \sqrt[3]{2} \\ \phi \in \{ 0, 2\pi/3, 4\pi/3 \} $$ This gives $$ z \in \{ \sqrt[3]{2}, \sqrt[3]{2} \, e^{i \cdot 2\pi/3}, \sqrt[3]{2} \, e^{i \cdot 4\pi/ 3} \} $$