Solve the equation for positive integers $x,y$
$$y^3+3y^2+3y=x^3+5x^2-19x+20$$
i have written the equation as
$$(y+1)^3=x^3+5x^2-19x+21$$ which is
$$(y+1)^3-x^3=5x^2-19x+21$$ and by using $a^3-b^3$ we get
$$(y+1-x)((y+1)^2+(y+1)x+x^2)=5x^2-19x+21$$
now since the discriminant of $5x^2-19x+21$ is negative, the quadratic is always positive, so $y+1 \gt x$
Now if RHS is a prime number then the only possibility is
$y+1-x=1$ i.e., $y=x$ and substituting $x=y$ in original equation we get $2x^2-22x+20=0$ $\implies$ $x^2-11x+10=0$
Hence $x=1,y=1$ and $x=10,y=10$ are Possible solutions.
But if RHS is not a prime number how can i proceed?
You did a great job that LHS is a perfect cube. Now your task is to figure out when is $x^3+5x^2-19x+21$ also a perfect cube. Notice that for $x>10$, you have $(x+1)^3 < x^3+5x^2-19x+21 < (x+2)^3$, therefore it can't be a perfect cube. So you have to check the cases when $x\in\{1,2,\ldots,10\}$.