How to solve system of equations? I want to sove the system of equations $$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}
$$
I tried divide these two equations we obtain
$$\dfrac{x^3 - y^3}{(x+y)^2 } = \dfrac{7}{9}$$
From here, I don't know how to solve.

$$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}$$
$x^3y-y^4=y(x^3-y^3)$, while $x^2y+2xy^2+y^3=y(x+y)^3$. 
We try the transformation $x+y=z$. 
Then $y(x^3-y^3)=y(x-y)(x^2+xy+y^2)=y(z-2y)(z^2-xy)=y(z-2y)(z^2-(z-y)y)=$
$-2y^4+3y^3z-3y^2z^2+yz^3=7$
On the other hand, the second equation gives $yz^2=9$, so $y=\frac{9}{z^2}$. We expand and then notice that it is actually equal to $9(z^3-18)(z^6-9z^3+81)=7z^8$. Then we get $z=3$ as a root. 
The problem is that I do not know how do I show that there are no other roots, but probably someone who knows calculus can be able to resolve this? or probably by a weird factoring?
