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 :)

Reading the comments, I now realize that from the equation $x^3 + \frac{1}{x^3} = 2\sqrt5$, if we let $z = x^3$, then we get $z + \frac{1}{z} = 2\sqrt5 \Longrightarrow z^2 + 1 - 2\sqrt5z = 0 \Longrightarrow$ using the quadratic formula, we get $z = \sqrt5 \pm 2$. Hence, from here, we can get that the answer is $5 - 2 = 3$.
I could have also tried different values of $y$ to get $\sqrt5$ as the answer.
