$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$ 
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
 x^2+xy+y^2=3 \end{cases}$$
  by reducing the system to a system of second degree. 

What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
 x(x+y)+y^2=3 \end{cases}$$
 A: Hint:
$$x^4+y^4+x^2y^2=(x^2+y^2)^2-(xy)^2=?$$
So, we know $x^2+y^2,xy=9>0$
So, $x,y$ will have the same sign
Hope you can take it  from here using $y=\dfrac9x$ 
A: You can substitute $u=x+y$ and $v=xy$. Then
$$x^2+xy+y^2=(x+y)^2-xy=u^2-v $$
and
$$x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(u^2-2v)^2-v^2 $$
You obtain the system
$$\begin{cases}
u^4-4u^2v+3v^2=21\\ u^2-v=3
\end{cases}$$
From the second equation you get $v=u^2-3$. Substitute that in the first equation, and you'll immediately find $u^2=1$. After that, you can find $u$ and $v$, and I assume you can finish.
A: Multiply the first by $x^2-y^2$ and the second by $x-y$.$$x^6-y^6=21(x^2-y^2),\\x^3-y^3=3(x-y)$$
Then take the ratio
$$x^3+y^3=7(x+y).$$
Adding the two above,
$$2x^3=10x+4y$$
and using $2y=x^3-5x$, 
$$4x^2+2x(x^3-5x)+(x^3-5x)^2-12=x^6-8x^4+19x^2-12=0.$$
By trial and error, $x^2=1$ are solutions and we factor as
$$(x^2-1)(x^4-7x^2+12)=(x^2-1)(x^2-3)(x^2-4).$$
A: I found $$9=(x^2+xy+y^2)^2=$$ $$=(x^4+x^2y^2+y^4)+2(x^2y^2+x^3y+xy^3)=$$ $$=21+2xy(xy+x^2+y^2)=21+2xy(3)=$$ $$=21+6xy$$ giving $xy=-2.$
Then $(x+y)^2=(x^2+xy+y^2)+xy=3+xy=3-2=1.$
Knowing  $(S,P)=(\pm 1,-2),$ we can find all possible $\{x,y\}$ as the solutions of $0=z^2-Sz+P=0=z^2\pm z-2.$
