### Instruction: If the sum of three real numbers is $0$ and their product is $17$, then what is the sum of their cubes?### Response: Let the three numbers be $a$, $b$, and $c$.
Then I have the system of equations $a + b + c = 0$ and $abc = 17$.
I want to find the sum of their cubes, $a^3 + b^3 + c^3$.
I wonder if I can use some identity or factorization to simplify this expression.
I recall that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
This looks promising, because I know the values of $a + b + c$ and $abc$.
Plugging them in, I get $a^3 + b^3 + c^3 - 3(17) = (0)(a^2 + b^2 + c^2 - ab - bc - ca)$.
Simplifying, I get $a^3 + b^3 + c^3 = 3(17) = 51$.