If $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, then find $x^2 + \frac{1}{x^2}$. Question:  If $x$ is a real number satisfying $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, determine the exact value of $x^2 + \frac{1}{x^2}$.
My partial solution: $(x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}.$
We know that $x^3 + \frac{1}{x^3} = 2\sqrt5 \Longrightarrow (x + \frac{1}{x})^3 = 2\sqrt5 + 3x + \frac{3}{x}.$
So, $(x + \frac{1}{x})^3 = 2\sqrt5 + 3(x + \frac{1}{x})$. Let $y = x + \frac{1}{x}$.
Then, $y^3 = 2\sqrt5 + 3y$. We want to solve for $x^2 + \frac{1}{x^2}$. $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$.
So, we want to solve for $y^2 - 2$. But, I'm confused about how to solve the equation $y^3 = 2\sqrt5 + 3y$, should I check random values?
Any help is appreciated, thanks in advance! (Thanks for the suggestions in the comments :)

From $x^3+\frac{1}{x^3}=2\sqrt{5}$ we get $\left(x^3+\frac{1}{x^3}\right)^2=20$ or $x^6+\frac{1}{x^6}=18$
Let $y=x^2+\frac{1}{x^2}$. Then $y^3 = 3y + x^6+\frac{1}{x^6} = 3y +18.$
The only real solution of $y^3-3y-18=0$ is $y=3$. Therefore $x^2+\frac{1}{x^2}=3.$
By the way, $x = \frac{\sqrt{5}\pm 1}{2}$ can be used to see that this is actually the solution.
