Given that: $$x^2 - xy + y^2 = 2$$ $$w^2 + wz + z^2 = 2$$ $$wz + 4 = xy$$
Evaluate $$\sqrt[5]{2(x^2z^2 + x^2y^2+4z^2)}$$
I tried to apply the Sophie Germain identity in the last expression, but failed. So, in order to get rid of $w$, I've manipulated those equations and didn't came to a conclusion so far. Can you help me with this seemingly simple algebra problem?
The first two equations can be expressed as:-
$$(x-y)^2+xy=2 \\(w+z)^2-wz=2$$ Summing these, and using the third equation, we have $$(x-y)^2=-(w+z)^2$$ which can only hold true if $x=y$ and $w=-z$, assuming all numbers are real.
This simply leads to $x^2=y^2=w^2=z^2=2$, so the result of your expression is $$\sqrt[5]{2(4+4+8)}=2$$