How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And 
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .

Denote $a=x/y$, $b=xy$.
Then
$$a+1+\frac{1}{a}=\frac{4}{b},$$
$$a^2+1+\frac{1}{a^2}=\frac{8}{b^2},$$
$$a^3+1+\frac{1}{a^3}=?$$
Then from the $1$st equation: 
$$
\left(a+\frac{1}{a}\right)^2 = \left(\frac{4}{b}-1\right)^2,
$$
combining it with $2$nd equation, $\Rightarrow$ $b=1$.
Then, $a+\frac{1}{a}=3$, $a^2+\frac{1}{a^2}=7$; therefore $a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right) - \left(a+\frac{1}{a}\right) = 21-3=18$.
Then $a^3+1+\frac{1}{a^3}=19$. Myltiplying by $b^3=1$, $\Rightarrow$ $x^6+x^3y^3+y^6=19$.
