Algebraic System

a) Note that $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$

Substituting what we have, we get $1 = 1 (a^2 + b^2 + ab)$, which means that $a^2 + b^2 + ab = 1$

We can write $a^2 + b^2 + ab$ as $(a - b)^2 + 3ab$. 

Using what we know, we have $1^2 + 3ab = 1$, meaning $ab = \color{brown}\boxed{0}$

b) We already proved that $a^2 + b^2 + ab = 1$.

Now, we write this as $(a+b)^2 - ab = 1$

Because $ab = 0$, we know $(a+b)^2 = 1$, meaning $a + b = \color{brown}\boxed{-1, 1}$ 

c) If ab = 0, either a or b = 0. 

If a = 0, b = -1, so $(a, b)$ is $\color{brown}\boxed{(0, -1)}$

If b = 0, a = 1, so $\color{brown}\boxed{(1, 0)}$

These are the only solutions that work