Other Heuristic Method to Prove the Given Expression Greater Than Zero

Given that $x\ne 0$ and $x$ is real that satisfies $\large \sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.

Regarding to the previous blog post(china-imo-contest-problem) that I mentioned how to prove

$a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x> 0$ in

$\large (a-x) ((a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x))=0$, my method is not the only way out.

I have seen some other good approach that I wish to share with you as well.

The other method recognized that $a^2+ax+\left(\dfrac{x}{k}\right)^2=\left(a+\dfrac{x}{2}\right)^2+kx^2+\dfrac{(4-k)x^2}{4k}> 0$ for $a,\,x\ne 0$ and $k\le 4$.

This turns the target expression into

$a^4+a^3 x+a^2 x^2+a x^3+x^4+x^2+a x+1$

$=\underbrace{a^2\left(a^2+a x+\dfrac{x^2}{4}\right)}+\underbrace{x^2\left(x^2+a x+\dfrac{a^2}{4}\right)}+\dfrac{a^2x^2}{2}+\underbrace{(x^2+a x+1)}$

$=a^2\left(\left(a+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4}\right)+x^2\left(\left(x+\dfrac{a}{2}\right)^2+\dfrac{a^2}{4}\right)+\left(\left(a+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4}\right)$

$> 0$ for $a,\,x\ne 0$ and $k\le 4$.

This is other workable method and as someone who is the strong advocate for teaching of heuristic mathematical problem solving skills, it's another method that I would like to ask the students to think about it for a moment and absorb the key point of it and 99% of the time it would be another useful weapon that we have got to solve for other intriguing math problem elegantly.