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 .

Note that 
$$(x^2+xy+y^2)^2-2xy(x^2+xy+y^2)=x^4+x^2y^2+y^4 $$
so that
$$ xy = 1.$$
Then 
$$\begin{align}x^6+x^3y^3+y^6&=(x^4+x^2y^2+y^4)(x+xy+y^2)-xy^5-2x^2y^4-2x^4y^2-yx^5\\
&=8\cdot 4-xy(x^4-2x^3y-2xy^3-y^4)\\
&=32-1\cdot ((x^4+x^2y^2+y^4)-xy(2x^2-xy-2y^2))\\
&=24+1\cdot(2(x^2+xy+y^2)-3xy)\\
&=29
\end{align} 
$$
(I suppose) Just keep subtracting simple products and powers of the given polynomials to get rid of moniomials not divisible by $xy$, then divide out $xy$ from the rest and continue.
Alternatively to the above, you might also start by subtracting $(x^2+xy+y^2)^3$, which would also "kill" the $x^6$ and $y^6$.
