Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can i solve this problem?
PLEASE i need some help and thanks for all!!

Notice
$$(x^2 + y^2)^2 = 64 \implies x^4 + y^4 + 2(xy)^2  = 64$$ 
and
$$ (x + y )^2 = 4 \implies x^2 + y^2 + 2xy = 4 \implies 2xy = 4 - 8 = -4 \implies xy = -2
$$
$$ \therefore x^4 + y^4 = 64 - 2(xy)^2 = 64 - 2(-2)^2 = 56  $$
