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:
  $x^3=1$.
  
  $\iff (x-1)(x^2+x+1)=0$
  
  $\Rightarrow x=1$ or
  $x^2+x+1=0$.
  Let
  $\mathrm{\Delta }=1^2-4\cdot 1\cdot 1=-3\Rightarrow \sqrt{\mathrm{\Delta }}=\sqrt{3}i$.
  
  $x=\frac{-1\pm \sqrt{3}i}{2}=\frac{-1}{2}\pm \frac{\sqrt{3}}{2}i$.
• Case 2:
  $x^3=2$.
  
  $\iff (x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+{\sqrt[3]{2}}^2)=0$
  
  $\Rightarrow x=\sqrt[3]{2}$ or
  $x^2+\sqrt[3]{2}x+{\sqrt[3]{2}}^2=0$.
  Let
  $\mathrm{\Delta }={\sqrt[3]{2}}^2-4\cdot 1\cdot {\sqrt[3]{2}}^2=-3{\sqrt[3]{2}}^2\Rightarrow \sqrt{\mathrm{\Delta }}=\sqrt[3]{2}\sqrt{3}i$.
  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:

$$S=\{-1,\sqrt[3]{2},\frac{-1}{2}\pm \frac{\sqrt{3}}{2}i,\frac{-\sqrt[3]{2}}{2}\pm \frac{\sqrt{3}}{\sqrt[3]{4}}i\}.$$