Computing the value of $(x+y)^4$ if $x^4+y^4=5$ and $x^2+xy+y^2=10$ 
Let $x^4+y^4=5$ and $x^2+xy+y^2=10.$ Find $(x+y)^4.$

First, I tried expanding $(x+y)^4$ using the binomial theorem to get $5+4x^3y+6x^2y^2+4xy^3,$ so simplifying I got $5+4xy(x^2+y^2)+6(xy)^2.$ Then I rearranged the given equation to get $x^2+y^2=10-xy,$ so the expansion becomes $5+4xy(10-xy)+6(xy)^2.$ I further simplified to get $2x^2y^2+40xy+5,$ but I'm not sure how to continue off here. Factoring this expression doesn't seem to help. May I have some help? Thanks in advance.

An alternative approach is to rearrange the second equality and square:
$$
x^2 + y^2 = 10 - xy \implies \underbrace{x^4 + y^4}_{=\, 5} + 2 x^2y^2 = 100 - 20 xy +x^2 y^2 \implies x^2y^2 + 20 xy = 95
$$
The last quadratic could be solved for $\,xy\,$, then $\,x,y\,$ determined and $\,(x+y)^4\,$ calculated by brute force. A shortcut, instead, is to again rearrange the second equality in a different way and square:
$$
(x+y)^2 = 10 + xy \implies (x+y)^4 = 100 + 20 xy + x^2y^2 = \dots
$$
