<s>[INST] 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.
 [/INST] Actually, there is only one solution in this equation.
let $f(x)=(x^2+4)^{1/3}-\sqrt{x-1}-2x-3(x\geq 1)$,then $f'(x)=\frac{2x}{3(x^2+4)^{\frac{2}{3}}}-\frac{1}{2\sqrt{x-1}}-2 \leq \frac{2x}{3(x^2+4)^{\frac{2}{3}}} -2\leq \frac{2x}{3(x^2)^{\frac{2}{3}}} -2=\frac{2}{3x^{\frac{1}{3}}}-2 \leq -\frac{4}{3}$.
So $f(x)$ is strictly decreasing function and this equation has only one solution.
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