<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] 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.
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