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 .

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, 
$$x^3 y^3 = 1, \; \; \; x^6 + y^6 = 18,  $$
$$ x^6 + x^3 y^3 + y^6 = 19. $$
