Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks

Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.

I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither could think of any other ways to proceed.
How to proceed?
Thank you.

You are asking for
$$  x^3 + 5 x^2 - 19 x + 21 = (y+1)^3 $$
For large enough positive $x,$ (you need to find out explicit lower  bound for $x$),
$$ (x+1)^3 < x^3 + 5 x^2 - 19 x + 21 < (x+2)^3  $$
and cannot be a cube. Then check the small values of $x$ remaining.
so, $x^3 + 3 x^2 + 3x + 1 < x^3 + 5 x^2 - 19 x + 21,$ or
$0 < 2 x^2 -22x + 22$ or $x^2 - 11 x + 11 > 0.$ This one is true for $x \geq 10,$ either draw a picture or...
The other one is $x^3 + 5 x^2 - 19 x + 21 < x^3 + 6 x^2 + 12 x + 8,$  or $0 < x^2 + 31 x - 13.$ This one is true for integers $x \geq 1.$
So, check the original problem for $x = 0,1,2,3,4,5,6,7,8,9.$
