China IMO Contest Problem: Given $\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.

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 $x$.

Answer:

Let $\large a=\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$.

Consider  $\large a=\sqrt[3]{x^5-20x}$, we have:

$\large a^3=x^5-20x---(1)$

Consider $\large a=\sqrt[5]{x^3+20x}$, we have:

$\large a^5=x^3+20x---(2)$

Adding the equations $1$ and $2$ up we get:

$\large a^3+a^5=x^5+x^3$

Collect terms with the same power of exponent, we obtain:

$\large a^3+a^5=x^5+x^3$

$\large a^5-x^5=x^3-a^3$

Factor the above equation by using the difference of cubes and fifth exponents we have:

$\large (a-x) (a^4+a^3 x+a^2 x^2+a x^3+x^4)=(x-a) (a^2+a x+x^2)$

$\large (a-x) (a^4+a^3 x+a^2 x^2+a x^3+x^4)-(x-a) (a^2+a x+x^2)=0$

$\large (a-x) (a^4+a^3 x+a^2 x^2+a x^3+x^4)+(a-x) (a^2+a x+x^2)=0$

$\large (a-x) ((a^4+a^3 x+a^2 x^2+a x^3+x^4)+(a^2+a x+x^2))=0$

$\large (a-x) ((a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x))=0$ (*)

The key to solve for this problem is to recognize and then prove that the second factor in the equation (*) above is always greater than zero so that equation (*) is zero when the first factor equals zero.

In order to prove that $a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x> 0$, we have to group the right terms together, and this may require a few trial and error before we get it right:

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

[MATH]\color{black}=\color{yellow}\bbox[5px,purple]{(a^4+a^2 (x^2+1)+x^4+x^2)}\color{black}+\color{yellow}\bbox[5px,green]{(a^3 x+a x^3+a x)}[/MATH]

Note that:

1.

Since  $x\ne 0$, and that $\large a=\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$, that implies $a\ne 0$ and therefore we have:

[MATH]\color{yellow}\bbox[5px,purple]{(a^4+a^2 (x^2+1)+x^4+x^2)> 0}[/MATH]

2.

For [MATH]\color{yellow}\bbox[5px,green]{(a^3 x+a x^3+a x)=ax(a^2+x^2+1)}[/MATH], if we can prove that $ax> 0$ (since $a^2+x^2+1> 0$ for all real $a,\,x$ where $a\ne 0$ and $x\ne 0$),


then [MATH]\color{black}=\color{yellow}\bbox[5px,purple]{(a^4+a^2 (x^2+1)+x^4+x^2)}\color{black}+\color{yellow}\bbox[5px,green]{(a^3 x+a x^3+a x)}[/MATH] is certainly greater than $0$.

Effective use of "resources" (what we're given and what we have been mentioned so far) is the responsibility of real students who are creative thinkers. Look deep into the given resources and make full use of them to aid in our problem solving journey.

From $\large a^3=x^5-20x---(1)$, we could algebraically manage it so we get:

$ a(a^2)=x(x^4-20)$

$ \dfrac{a}{x}(a^2)=x^4-20---(3)$

and to manage $\large a^5=x^3+20x---(2)$ in the similar fashion to get:

$ a(a^4)=x(x^2+20)$

$ \dfrac{a}{x}(a^4)=x^2+20---(4)$

Adding both equations (3) and (4) gives:

$\dfrac{a}{x}(a^2)+ \dfrac{a}{x}(a^4)=x^4-20+x^2+20$

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

$\dfrac{a}{x}=\dfrac{x^2(x^2+1)}{a^2+a^4}> 0$

Therefore, observations from points (1) and (2) suggest that $a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x> 0$.

Looking back at the equation (*) where

$\large (a-x) ((a^4+a^3 x+a^2 (x^2+1)+a x^3+x^4+x^2+a x))=0$ (*)

It equals zero if and only if $a-x=0$, i.e. $a=x$.

So we have

$x^3=x^5-20x$

$x^5-x^3-20x=0$

$x(x^4-x^2-20)=0$

$x(x^2-5)(x^2+4)=0$

The only real values of $x$ that satisfy the above equation are $-\sqrt{5},\,\sqrt{5}$ and hence their product is $5$ and we're done.