Solving $\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$ I’m having a hard time with this radical equation.
$$\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$$
I subtracted square root of x^2-3 then I got ^3 square root of -x+4 + square root of x^2-3 = 1. Next I cube both sides which I got -x^2-3= (1- square root of x^2-3) ^3. Would o need to add that back to the other side and use difference of cubes ?

HINT
We can let $u = x^{2} - 3$, where $x^{2} - 3 \geq 0$. Then we get that
\begin{align*}
\sqrt[3]{1 - u} + \sqrt{u} = 1 & \Longleftrightarrow \sqrt[3]{1-u} = 1 - \sqrt{u}\\\\
& \Longleftrightarrow 1 - u  = 1 - 3\sqrt{u} + 3u - \sqrt{u^{3}}\\\\
& \Longleftrightarrow \sqrt{u^{3}} - 4u + 3\sqrt{u}  =0\\\\
& \Longleftrightarrow (u - 4\sqrt{u} +3)\sqrt{u} = 0 
\end{align*}
Can you take it from here?
