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.
 A: 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 $x = 0,1,2,3,4,5,6,7,8,9.$
A: You can rewrite it as $$(y+1)^3 = x^3 + 5x^2 - 19x + 21$$
Now $y+1$ and $x$ differ for some integer, say $z$, so $(y+1) - x=z$, so we have $$(x+z)^3 =  x^3 + 5x^2 - 19x + 21$$ or $$ x^2(3z-5)+x(3z^2+19)+(z^3-21)=0$$ so we have a quadratic equation on $x$ with an integer parameter $z$: It will have an integer solution if it discriminat is a perfect square, so
$$(3z^2+19)^2-4(3z-5)(z^3-21) = d^2$$ or $$-3z^4+20z^3+114z^2+252z-59 =d^2$$
Notice that $$-3z^4+20z^3+114z^2+252z-59\geq 0$$ but that can not happend very often.
A: 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 $K(10,10)=1000+300+30-(1000+500-190+20)=0$.
