If $x+y=1$ and $x^{2}+y^{2}=2$, calculate $x^{3}+y^{3}$.
We have the binomial identity, $(x+y)^{2}=x^{2}+2 x y+y^{2}$, and the trinomial identity, $(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$. Substituting the values of $x+y$ and $x^{2}+y^{2}$ into the binomial identity, we get $1=2+2 x y$ and, therefore, $x y=-\frac{1}{2}$. Thus, by the trinomial identity,

$$
x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)=1-3 \cdot\left(-\frac{1}{2}\right) \cdot 1=\frac{5}{2}
$$