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.
 A: $$\sqrt[3]{x^2+4}=\sqrt{x-1}+2x-3$$
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 $x=2$.
A: 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.
A: If you don't have calculus available but know about continuous functions I would reason as follows.  The equation doesn't make sense when $x \lt 1$ because the square root on the right doesn't work.  At $x=1$ the left side is positive and the right is negative.  As $x$ gets very large the right is about $2x$ and the right is $x^{2/3}$ so the right will be greater.  There must be at least one root with $x \gt 1$.  Now I would try small integral values for $x$, both hoping to get lucky and if not finding where the sides cross.  Bingo!  $x=2$ works.  If it didn't, say because the constant on the right was $-4$ instead of $-3$ you would know there was a root between $x=2$ and $x=3$ and you could use bisection to locate it.  
This is an example of my personal rational root theorem.  For class problems the root is often an integer less than $20$.  If not an integer it is rational with numerator and denominator less than or equal to $10$.  That doesn't leave many to try.
A: 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 $x=2$.

