# 2 + 4 solutions

## Problem statement
Find all possible solutions to the equation:
$$x^6 - x^3 = 2.$$

## Solution
Rewrite the equation as follows: $(x^3)^2 - x^3 - 2 = 0$.
- Case 1: $x^3 = 1$.
$\Leftrightarrow (x - 1)(x^2 + x + 1) = 0$
$\Rightarrow x = 1$ or $x^2 + x + 1 = 0$.
Let $\Delta = 1^2 - 4 \cdot 1 \cdot 1 = -3 \Rightarrow \sqrt{\Delta} = \sqrt{3}i$.
$x = \frac{-1 \pm \sqrt{3}i}{2} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2}i$.
- Case 2: $x^3 = 2$.
$\Leftrightarrow (x - \sqrt[3]{2})(x^2 + \sqrt[3]{2}x + \sqrt[3]{2}^2) = 0$
$\Rightarrow x = \sqrt[3]{2}$ or $x^2 + \sqrt[3]{2}x + \sqrt[3]{2}^2 = 0$.
Let $\Delta = \sqrt[3]{2}^2 - 4 \cdot 1 \cdot \sqrt[3]{2}^2 = -3\sqrt[3]{2}^2 \Rightarrow \sqrt{\Delta} = \sqrt[3]{2}\sqrt{3}i$.
Therefore, $x = \frac{-\sqrt[3]{2} \pm \sqrt[3]{2}\sqrt{3}i}{2} = \frac{-\sqrt[3]{2}}{2} \pm \frac{\sqrt{3}}{\sqrt[3]{4}}i$.

Finally, the set of possible solutions to the given equation can be represented below:
$$S = \{-1, \sqrt[3]{2}, \frac{-1}{2} \pm \frac{\sqrt{3}}{2}i, \frac{-\sqrt[3]{2}}{2} \pm \frac{\sqrt{3}}{\sqrt[3]{4}}i\}.$$