Let $S$ be the set of rational numbers $x$ such that $x^3-432$ is the square of a rational number.

If $S$ is infinite, enter $-1$ . If $S$ is finite, enter the sum of the elements of $S.$

**
Here is a very helpful hint:
**

If
$y^2=x^3-432,$
let
$u = \dfrac{36-y}{6x}, \, v = \dfrac{36+y}{6x}$
. What is
$u^3+v^3$
?

The answer is 12.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Using the

very helpful hint(thanks, Patrick!), let$u = \dfrac{36-y}{6x}, \, v = \dfrac{36+y}{6x}$

Then

$u^3+v^3=\dfrac{432+{y}^{2}}{{x}^{3}}$

so that we immediately have, as a consequence

$u^3+v^3=1$

which has no known rational solutions per Fermat's Last Theorem , other than

$(u,v)=(1,0)$ or $(u,v)=(0,1)$

which leads to the only rational solutions

$(x,y)=(12,-36)$ or $(x,y)=(12,36)$

and so the sum of the elements of $S$ is just $12$