<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] 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
$$(x^2-1)(x^4-7x^2+12)=(x^2-1)(x^2-3)(x^2-4).$$
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