finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

$x^2+y^2=(x+y)^2-2xy=16-4=12.$
$x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$.
Now you can use your factorization:
$x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.
