### Instruction: If $x+y=4$ and $x^2+y^2=8$, find $x^3+y^3$.### Response: 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 $x^3+y^3=(x+y)(x^2-xy+y^2)$.