TITLE: How to solve this equation manually: $(x^2+100)^2=(x^3-100)^3$?
QUESTION [7 upvotes]: Well, I was given a problem,
find $x$, if:
$$(x^2+100)^2=(x^3-100)^3$$
I tried everything that I could, I even opened up the brackets which gave an ugly degree 9 equation, I also tried to plot the curves $y=\left(x^2+100\right)^2$ and $y=\left(x^3-100\right)^3$ and locate their point of intersection but it couldn't be done manually.
So, in the end I was forced to use hit and trial after doing which I got the answer, is their any way to solve this algebraically??

REPLY [20 votes]: It's obvious that $x^3>100$ so $x>0$ .
Consider the $6$-th root of the equation to get :
$$\sqrt[3]{x^2+100}=\sqrt{x^3-100}$$
Now consider the function :$$f(x)=\sqrt[3]{x^2+100}$$
This function is bijective from $(0,\infty$) to $(\sqrt[3]{100},\infty)$ .
Its inverse is :$$f^{-1}(x)=\sqrt{x^3-100}$$
This means that the equation is now : 
$$f(x)=f^{-1}(x)$$
$$f(f(x))=x$$
But $f$ is an increasing function so let's take two cases :

If $f(x)>x$ then :

$$x=f(f(x))>f(x)>x$$ a contradiction .

If $f(x)<x$ then :
$$x=f(f(x))<f(x)<x$$ a contradiction .

This means that $f(x)=x$ so : $$x^3=100+x^2$$ which can be solved easily .