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.

Given:
 $$y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$$
Let $K(x,y)$ be:
$$K(x,y)=y^3 + 3y^2 + 3y - (x^3 + 5x^2 - 19x + 20)$$
We could write this as:
$$K(x,y)=f(y)-f(x)+g(x)$$
If a root $r$ exists for $g$, then, we would have:
$$K(r,r)=f(r)-f(r)+g(r) $$
We can find g(x) such that:
$$g(x)=K(x,y)-f(y)+f(x) \tag1$$
By inspecting K(x,y), we need to choose the function $f$ first, then calculate $g$ after that.
Let $f(y)$ be:
$$f(y)=y^3 + 3y^2 + 3y$$
This implies $f(x)$ be:
$$f(x)= x^3 + 3x^2 + 3x$$
Using (1) above:
$$g(x)=[y^3 + 3y^2 + 3y - (x^3 + 5x^2 - 19x + 20)] - (y^3 + 3y^2 + 3y) - (x^3 + 3x^2 + 3x)$$
Simplifying, we get:
$$g(x)=2x^2-22x+20$$.
The above equation has 2 real roots $r=1,r2=10$.
Testing these values, $K(1,1)=1+3+3-(1+5-19+20)=0$ and $K(10,10)=1000+300+30-(1000+500-190+20)=0$.
