Solving $(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$ I'm trying to solve the equation below for $x$: $$(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$$
I tried to solve it by doing some algebraic operations such as expanding both sides and then try to remove the third roots by rising both sides of the equation to power of 3 but it gets somehow complicated.
Someone told me that this equation can be solved by using derivation but I have no idea how to use derivation for solving this equation.
Any help would be appreciated.

As mentioned by @Player0, the equation can be solved using monotonicity noticing that $x=2$ is a solution. But in this answer, I will be solving the equation algebraically (and for real numbers).

We first remove the bizarre cube roots from our equation. To do that, we consider $y=\sqrt{x^3-4}$ and $z=\sqrt[3]{x^2+4}$ where $y,z>0$. So, we can rewrite our equation as $$(y^2)^3=(z^2+4)^2$$ $$\Rightarrow y^3=z^2+4$$ which looks good.
We also have the following systems of equations from our definition of $y$ and $z$ and from the above equation $$\left\{\begin{matrix}x^3-y^2=4\\ y^3-z^2=4\\ z^3-x^2=4\end{matrix}\right.$$ which are really beautiful.
Now in our original equation, since RHS is a square, we have $x^3-4 \geq 0 \implies x\geq\sqrt[3]{4}$. Now, we assume $x \geq z$. From the above systems of equations we have $$\left\{\begin{matrix}x^3-z^3=y^2-x^2\\ y^3-x^3=z^2-y^2 \end{matrix}\right.$$
Now since we had $x \geq z$, we have $y\geq x$ and $z \geq y$. Notice that we have $z\geq y\geq x$ which is not possible unless we have $x=y=z$. (Similarly contradiction happens if $x<z$).
Now, plugging in the values we have,
$x=\sqrt{x^3-4} \Rightarrow x^3-x^2-4=0 \Rightarrow (x-2)(x^2+x+2)=0$
This implies $\boxed{x=2}$ which is the only solution to the problem.
Remark: I think it is quite amazing that this irrational equation can be written as a beautiful system of equations.
