Solve for x and y, $x^3+7x^2+35x+27=y^3$ where x and y are integers.

Question

Okay, I have done parity check. I have taken cases of $x$ being odd or even and been working with mod 7 but nothing happened. How to proceed with further? Please help! This is my try In case 1, where $x$ is odd and $y$ is even $x^3+7x^2+35x+27≡0,5,6\bmod 7$ so $y^3≡6\bmod 7$ In case 2, where $x$ is even and $y$ is odd similarly we can show $y^3≡6\bmod 7$. Apologies, if I am doing anything wrong since I am a beginner.

Answer 1

Hint: $(x+2)^3<x^3+7x^2+35x+27<(x+3)^3$ outside some interval $[a,b]$. Determine possible values for $a,b$, prove that there cannot be any solutions $(x,y)$ where $x$ is outside this interval and check the finitely many cases in $[a,b]$.