Irrational equation, How to solve it? The equation 
$$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ has a solution $x = 3.$ How to solve this eqution?
 A: The domain of the equation $$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ for real $x$ comes from the squareroot, so that it is $[2^{1/3},\infty)$. Thus we may safely divide through by $x$ and write it in the form $f(x)=g(x)$ where
$$f(x)=(x^{-1}+x^{-3})^{1/3}+1,\\ g(x)=(x-2x^{-2})^{1/2}.$$
One effect of the division by $x$ is that it makes it easier to analyze where the sides $f,g$ are increasing and decreasing. 
A simple calculation then shows $f$ is increasing on $[2^{1/3},\sqrt{3}]$ and thereafter decreasing on $[\sqrt{3},\infty)$, and that $g$ is increasing on the entire domain $[2^{1/3},\infty).$
Between $2^{1/3}$ and $\sqrt{3}$, $g$ increases from $0$ to about $1.03217$, while in the same interval $f$ increases from about $1.6647$ to $1.7247$. This means $f$ lies completely above $g$ on the interval $[2^{1/3},\sqrt{3}]$ so that $f=g$ has no roots there.
After that, on $[\sqrt{3}, \infty)$, $f$ decreases while $g$ increases, so that there can only be one root in all of $f=g$, which as noted in the post is at $x=3$ where $f=g=5/3.$
[note that the approximated numbers have in fact rather simple exact expressions; just didn't want to clutter things up with that detail.] I'm not really that satisfied with a derivative approach, but at least after divisionj by $x$ it became tractable.
A: You have to rewrite this as
$$
  \sqrt[3]{x^2-1}= \sqrt{x^3-2}-x
$$
and take the 3rd power of both members obtaining
$$
   x^2-1=(x^3-2)\sqrt{x^3-2}-3x(x^3-2)+3x^2\sqrt{x^3-2}-x^3.
$$
Now, you can isolate the square root obtaining
$$
  3x^4+x^3+x^2-6x-1=(x^3+3x^2-2)\sqrt{x^3-2}.
$$
I think from here you can go on by yourself.
