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 .
 A: 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 $y^6$.
A: why not.
$$  x^4 + x^2 y^2 + y^4 = (x^2 + xy + y^2)(x^2 - xy+y^2). $$
Worth memorizing. So
$$ x^2 - xy + y^2 = 2.  $$ Also
$$ 2xy = 2, \; \; \; xy=1. $$
And
$$ x^2 + y^2 = 3. $$
$$ 27 = (x^2 + y^2)^3 = x^6 + 3 x^4 y^2 + 3 x^2 y^4 + y^6 = x^6 + y^6 + 3 x^2 y^2 (x^2 + y^2) = x^6 + y^6 + 3 \cdot 1 \cdot 3 = x^6 + y^6 + 9  $$
So, 
$$x^3 y^3 = 1, \; \; \; x^6 + y^6 = 18,  $$
$$ x^6 + x^3 y^3 + y^6 = 19. $$
A: 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 $b^3=1$, $\Rightarrow$ $x^6+x^3y^3+y^6=19$.
