If $x+y=4$ and $x^2+y^2=8$, find $x^3+y^3$.
To solve this problem, I need to find a way to relate $x^3+y^3$ to the given equations. I notice that the first equation can be squared to get $(x+y)^2=16$, which expands to $x^2+2xy+y^2=16$. Comparing this with the second equation, I see that $2xy=16-8=8$, so $xy=4$. Now I wonder if I can factor $x^3+y^3$ using a sum of cubes identity. I recall that $x^3+y^3=(x+y)(x^2-xy+y^2)$. Plugging in the values of $x+y$ and $xy$, I get $x^3+y^3=4(8-4)=4(4)=16$.

