Show that $ x^3 - 6 = 25y^2 + 35y $ doesn't have any non-zero integer solution.
What I have tried
$$ 25y^2 + 35y - x^3 + 6 = 0 $$ solving for y we get $$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$ so I need to show that $$ 25 + 4x^3= z^2 $$ doesn't have a solution or if it does I need to show that the fraction is not equal to an integer but I can't proceed.
On
$x^3-6=25y^2+35y$ can be transformed to $v^2=u^3+400$ with $x=u/4$ and $y=(v-28)/40$.
The $(u,v)$ equation is an elliptic curve which has $2$ finite torsion points at $(0, \pm 20)$, giving $x=0,y=-1/5$ and $x=0,y=-6/5$ as solutions.
The rank of the curve is $0$ (from standard software) so there are NO other rational points on the curve.
Thus, there are no integral solutions.
Alternative solution
Suppose $x^3 = 25y^2 + 35y + 6 = (5y+6)(5y+1).$
If $(5y+6)$ and $(5y+1)$ have a common factor, then this common factor must divide $5$. If the common factor is $> 1$, then it must be $5$. But this would imply that $5$ is a factor of $x^3$, which would imply that $5 | x^3$ which would imply that $5|(5y+6)(5y+1)$.
This is impossible. Therefore $(5y+6)$ and $(5y+1)$ are relatively prime. Therefore $(5y+6)$ and $(5y+1)$ must each be a perfect cube. This is impossible, because for any positive integer $n$,
$[(n+1)^3 - n^3] = 3n^2 + 3n + 1 > 5.$
Note: It is therefore trivial that for $(5y+6)$ and $(5y+1)$ any perfect cubes, $(a^3)$ and $(b^3),$ where each of $(a^3)$ and $(b^3)$ may be positive or negative, you can't have $|(a^3) - (b^3)| = 5.$