How to solve this equation manually: $(x^2+100)^2=(x^3-100)^3$? 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??
 A: For a particular solution, you can take
$$x=5.$$
How did I find it? I restrict to the case of integers. We are looking for numbers such that $n^2=m^3$. Then, we expect to have $n=k^3$ and $m=k^2$ for some integer $k$. Now, we look for
$$k^3=x^2+100,\quad k^2=x^3-100.$$
Those two equations would be satisfied if $x=k$, and
$$k^3=k^2+100.$$
Now this equation is much simpler. I thought of what integer $k$ is such that $k^3$ is a bit above $100$ and $5^3=125$ was an obvious candidate.
A: There is a quite dirty trick you can use here. Define
$$
f(x,t)=(x^2+t)^2-(x^3-t)^3\ .
$$
Then solve $f(x,t)=0$ for $t$ (not for $x$). There is a simple solution of the cubic equation in $t$
$$
t=x^2(x-1)\ .
$$
Then set $t=100$ and solve the cubic equation for $x$, yielding $x=5$ as the only real root.
A: 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 .
