Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$ 
Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$ for $x,y,z \in \mathbb{Z}$.

We must have $x^3+y^3 = 1006d$ and $x^3+z^3 = 1001d$ where $d$ is an integer. This means that $x^3+y^3 \equiv 0 \pmod{1006}$ and $x^3+z^3 \equiv 0 \pmod{1001}$. Rearranging the equation we also get $1001x^3+1001y^3 = 1006x^3+1006z^3$ so $$1001y^3 = 5x^3+1006z^3.$$ How can we continue?

Well, at least we have some solutions, like
$$\frac{669^3 + 337^3}{669^3 + 332^3} = \frac{1006}{1001}$$
This should be studied with elliptic curves. 
May assume $x$, $y$, $z$, rationals, and then may even assume $x=1$. We get
$$\frac{y^3+1}{z^3+1} = \frac{1006}{1001}\\
z^3 +1 = \frac{1001}{1006}(y^3 + 1)$$
This is an elliptic curve, and it has an (easy) rational point ( does not give yet a solution to the problem) $(-1,-1)$. Now, take the tangent to the curve at this point and consider the intersection with the curve. It will be another rational point $(y,z) = (\frac{337}{669}, \frac{332}{669})$. 
I lack expertise in elliptic curves, so  will leave it here.
$\bf{Added:}$ Took the tangent line to the curve at the last point and intersected it with the curve. Got another point
$$(y,z) = (-11901775977431/50258598213909 , -13216294942936/50258598213909) $$
I guess the question now is whether our elliptic curve has finitely many rational points. 
