Find all solutions in N of the following Diophantine equation $(x^2 − y^2)z − y^3 = 0$
i divide by $z^3$ and look for rational solutions of the equation $A^2 − B^2 − B^3 = 0.$  The point $(A,B) = (0, 0)$ is a singular point, that is any line
through this point will meet the curve twice in $(0, 0)$. Now i wanna use Diophantus chord method using the lines
passing through $(0, 0)$ but i can't seem to pass this point

$$(x^2-y^2)z-y^3=0$$
$$ x^2 z = y^2 (y + z) $$
Let $λ$ be the gcd of $x$ and $y$, then we have:
$$ x = λn, y=λm, (m, n) = 1$$ 
and the equation becomes:
$$ n^2 z = m^2 (λ m + z) $$
Then, because (m,n)=1, we must have: $z=m^2 ω$
$$ n^2 ω = (λ m + m^2 ω) $$
$$ n^2 ω = m (λ + m ω) $$
and again, because (m,n)=1, it must be: ω = m k
$$ n^2 k = (λ + m^2 k) $$
$$ (n^2 - m^2) k = λ $$
with the only restriction being: (m,n)=1
So, all the solutions are of the form:
$$ x = k n (n^2 - m^2) $$
  $$ y = k m (n^2 - m^2) $$
  $$ z = k m^3 $$ 
