Let a and b be real numbers such that a^3 + 3ab^2 = 679 and 3a^2 b + b^3 = -679. Find a + b.
\(\begin{cases}a^3 + 3ab^2 = 679\\3a^2 b + b^3 = -679\end{cases}\)
Adding gives
\(a^3 + 3a^2 b + 3ab^2 + b^3 = 0\\ (a + b)^3 = 0\)
So a + b = 0.
