<s>[INST] $\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$ 
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:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
 x(x+y)+y^2=3 \end{cases}$$
 [/INST] 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 $0=z^2-Sz+P=0=z^2\pm z-2.$
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