We have: $2 = x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1 - 3xy$ Then: $xy = \dfrac{- 1}{3}$ So: $x^2 + y^2 = (x + y)^2 - 2xy = 1 - 2 . \dfrac{- 1}{3} = \dfrac{5}{3}$ and $x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = \dfrac{25}{9} - 2 . \dfrac{1}{9} = \dfrac{23}{9}$

$x+y = 1 \rightarrow 1 = (x+y)^2 = x^2 + 2xy + y^2 $ $ 2= x^3 + y^3 = (x+y)(x^2-xy+y^2) = x^2-xy+y^2 $ So we can find $xy = \frac{-1}{3}$. Easily we can find $x^2 + y^2 = 1 - 2xy = \frac{5}{3}$. just use this formula to find $x^4 + y^4 = (x^2+y^2)^2 - 2(xy)^2$

SOLUTION $1=(x+y)^3=x^3+3x^2y+3xy^2+y^3=$ $2+3xy(x+y)+2+3xy$, so $xy=-\frac{1}{3}$. Similarly, $1=(x+y)^2=$ $x^2+2xy+y^2=x^2+y^2-\frac{2}{3}$, so $x^2+y^2=\boxed{\frac{5}{3}}$. Lastly, $\frac{25}{9}=(x^2+y^2)^2=$ $x^4+2x^2y^2+y^4=x^4+\frac{2}{9}+y^4$, so $x^4+y^4=\boxed{\frac{23}{9}}$.