Let $S = a + b$ and $P = ab.$ By the given equation $a - b = 1,$ we can substitute this in as follows to express $a$ and $b$ in terms of $S$ and $P:$ \begin{align*}
a &= \frac{1}{2}\left((a + b) + (a - b)\right) = \frac{1}{2}\left(S + 1\right), \quad \text{and} \\
b &= \frac{1}{2}\left((a + b) - (a - b)\right) = \frac{1}{2}\left(S - 1\right).
\end{align*}Substituting these expressions into $a^3 - b^3 = 1$ to get rid of $a$ and $b,$ we have
\begin{alignat*}{8}1 \quad &= \quad&& a^3 &&\quad - \quad&& b^3 \\&= \quad&&\left(\frac{1}{2}\left(S + 1\right)\right)^3 &&\quad - \quad&& \left(\frac{1}{2}\left(S - 1\right)\right)^3 \\&= \quad&& \frac{1}{8}\left(S + 1\right)^3 &&\quad - \quad&& \frac{1}{8}\left(S - 1\right)^3 \\&= \quad&& \left(S + 1\right)\left(S^2 - S + 1\right) &&\quad - \quad&& \left(S - 1\right)\left(S^2 + S + 1\right) \\&= \quad&& (S^2 + S + 1) &&\quad - \quad&& (S^2 - S + 1) \\&= \quad&& (2S) &&\quad.
\end{alignat*}Thus, $2S = 1.$ Therefore, $S = \frac{1}{2}$ and $P = S - 1 = -\frac{1}{2}.$ Therefore, the possible values of $ab$ are $\boxed{-\frac{1}{2}},$ the possible values of $a + b$ are $\left\{\frac{1}{2}\right\},$ and the possible values of $a$ and $b$ are $\left\{\frac{1}{2}, -1\right\}.$
The real numbers $a$ and $b$ satisfy $a - b = 1$ and $a^3 - b^3 = 1.$
(a) Find all possible values of $ab.$
(b) Find all possible values of $a + b.$
(c) Find all possible values of $a$ and $b.$
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