How to solve these equations for x and y.. equations are 

$(x-y)(x+2y)(2x+y) = 20$ 

and

$x^2+xy+y^2 = 7$

i want the METHOD not the solutions 

Drawing the two associated curves, we see that there are $6$ simple solutions (the maximum according to Bezout theorem). Let $s=x+y,p=xy$. If $(x,y)$ is a solution, then $(-y,-x)$ is also a solution. Thus the solutions in $p$ have multiplicity $2$ and $p$ is any root of a polynomial of degree $3$. In the same way, the solutions in $s$ are pairwise opposite.
We obtain $s^2=p+7$ and $(x-y)(2s^2+p)=2$ or $(x-y)(3p+14)=20$ that implies $(s^2-4p)(3p+14)^2=400$ or $(7-3p)(3p+14)^2=400$. The solutions are $p=-6,-3,2$ ; the corresponding values of $s$ are $\pm 1,\pm 2,\pm 3$. The associated solutions in $(x,y)$ are $(-2,3),(-3,2)$ and $(3,-1),(1,-3)$ and $(-1,-2),(2,1)$.
