<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] As you gave,the part is $$\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}-2=0$$
$$\implies\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}=2..........(1)$$
$${\text{Let,}}$$
$$f(x)=\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}-\dfrac{1}{\sqrt{x-1}+1}$$
I will break $f(x)$ in two parts
$$g(x)=\dfrac{x+2}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}$$
$$h(x)=\dfrac{1}{\sqrt{x-1}+1}$$
Now if equation (1) is valid then $g(x)-h(x)$ is must be $2$.
I will check is it possible or not.I will start with h(x).
For h(x) to be valid it is sure that $x \geq 1$ and for any value of $x, 0\lt h(x)\leq 1$
Now in $g(x)$ it is clear that for $x\geq 1 ,numerator\lt denominator$.So,it is also true that,$0\lt g(x)\lt 1$
 
$$Hence, g(x)-h(x) \neq 2$$
$$\implies f(x)\neq 2$$
so,there is no solution for $f(x)$ and there exists only one solution and that is $x=2$.

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