How to solve this equation manually: $(x^2+100)^2=(x^3-100)^3$?

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.

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 .

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.