algebra precalculus - Solve the equation $sqrt[3]{x^{2}+4}=sqrt{x-1}+2x-3$

Solve the equation:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$
Things I have done so far:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$ Deducting 2 from both sides of equation
$$\Leftrightarrow (\sqrt[3]{x^2+4}-2)=(\sqrt{x-1}-1)+2x-4$$
$$\Leftrightarrow \frac{x^2+4-8}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$$
$$\Leftrightarrow (x-2)(\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2)=0$$
+)$$x-2=0\Leftrightarrow x=2$$
+)$$\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2=0(*)$$

I don't know how to solve the equation (*).
By the way, I think there must be "smart" way to solve this equation.

Answer

$$\sqrt[3]{x^2+4}=\sqrt{x-1}+2x-3$$

Let $x=t^2+1$ with $t \ge 0$. Then

\begin{align}
\sqrt[3]{t^4+2t^2+5}-t &= 2t^2-1 \\
\sqrt[3]{t^4+2t^2+5} &= 2t^2+t-1 \\

t^4 + 2t^2 + 5 &= 8t^6 + 12t^5 - 6t^4 - 11t^3 + 3t^2 + 3t - 1 \\
8t^6 + 12t^5 - 7t^4 - 11t^3 + t^2 + 3t - 6 &= 0 \\
(8t^5 + 20t^4 + 13t^3 + 2t^2 + 3t + 6)(t - 1) &= 0 \\
\end{align}

Note, for $t \ge 0$, that $8t^5 + 20t^4 + 13t^3 + 2t^2 + 3t + 6 > 0$. It follows that the only positive solution is $t=1$. Hence $x=2$.