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.
 A: We'll prove that $(2,1)$ is the only real solution.  First note that if $y = 0$, the second equation is unsolvable, so $y\neq 0$.
Solving the second equation for $x$ via the quadratic formula gives \begin{align*} x &= \frac{-2y^2 \pm \sqrt{4y^4-4y(y^3-9)}}{2y}\\ &= \frac{-y^2\pm 3\sqrt{y}}{y} \\ &= -y \pm y^{-\frac{1}{2}}.\end{align*}
In particular, $y \geq 0$.  Subbing this into the first equation gives \begin{align*}7 &= \left(-y \pm 3y^{-\frac{1}{2}} \right)^3 y - y^4 \\ &=  -2y^4\pm 9y^{\frac{5}{2}}-27y\pm 27y^{-\frac{1}{2}}.\end{align*}
This is equivalent to $$ 0 = -2y^{\frac{9}{2}} \pm 9y^{\frac{6}{2}}-27y^{\frac{3}{2}} -7y^\frac{1}{2} \pm 27.$$
Now, substitute in $z = \sqrt{y}$, which is possible since we already know $y\geq 0$.  This gives $$0 = -2z^9 \pm 9z^6 - 27z^3 -7z \pm 27$$
If we choose the $-$ sign, then clearly there is no solution with $z$ real, so we must have $$0 = -2z^9 + 9z^6 - 27z^3-7z+27.$$
Now, using the rational roots theorem, maple, direct inspection, etc, we see $z=1$ is a solution.
I claim that $f(z)-2z^9 + 9z^6 - 27z^3 - 7z + 27$ is always decreasing, so $z=1$ is the only solution.
To see this, compute the derivative, getting $f'(z)=-18z^8 + 54z^5-81z^2 - 7 = 9(-2z^8 + 6z^5-9z^2 - \frac{7}{9})$.  I claim this is always negative.  The $9$ out in front doesn't effect this, so we'll ignore it in intermediate computations.
Why is $f'(z)$ always negative?  Lets find its maximum.  So, take another derivative, getting $f''(z)-16z^7+30z^4-18z = 2z(-8z^6 + 15z^3 - 9)$.  Of course, this is $0$ when $z=0$, but the other part is quadratic in $z^3$ and so we see that it's never $0$.  Since $f'(z)$ goes to $-\infty$ as $x\rightarrow \pm \infty$, it follows that if it only has one critical point, this must be a max.  So, the max of $f'(z)$ occurs when $z = 0$, where $f'(0) = -7 < 0$.
Since $f'(z) < 0$, $f(z)$ is always decreasing, so $z= 1$ must be its only solution.
When $z=1$, $\sqrt{y} = 1$ so $y=1$, and subbing back in shows that $x=2$.
A: Guess and check works: $x=2$, $y=1$. (I tried forcing $x+y$ to be 3 and that's what I got.)
For a more general method for finding all solutions, I suppose applying the appropriate root-finding algorithm to approximate the solutions could work.
A: $$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}$$
$x^3y-y^4=y(x^3-y^3)$, 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?
