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 $b.$
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 $\left\{\frac{1}{2}, -1\right\}.$
To find the possible values of $ab$, we can use the given information:
From the equation $a - b = 1$, we can solve for $a$ in terms of $b$ as $a = b + 1$.
Substituting this into the second equation $a^3 - b^3 = 1$, we have:
$(b+1)^3 - b^3 = 1$
Expanding and simplifying, we get:
$b^3 + 3b^2 + 3b + 1 - b^3 = 1$
Simplifying further, we have:
$3b^2 + 3b = 0$
Factoring out $3b$, we get:
$3b(b+1) = 0$
This gives us two possible solutions: $b = 0$ or $b = -1$.
(a) Therefore, the possible values of $ab$ are $0$ or $-1$.
Now let's find the possible values of $a + b$:
Using the equation $a - b = 1$, we can rewrite it as $a = b + 1$.
Substituting this into the expression $a + b$, we have:
$(b + 1) + b = 2b + 1$
So the possible values of $a + b$ are $2b + 1$.
Plugging in the possible values of $b$ from earlier, we get:
For $b = 0$, $a + b = 2(0) + 1 = 1$.
For $b = -1$, $a + b = 2(-1) + 1 = -1$.
(b) Therefore, the possible values of $a + b$ are $1$ or $-1$.
To find the possible values of $a$ and $b$, we can use the values we found for $a + b$ and solve for each variable.
For $a + b = 1$, using the equation $a - b = 1$, we get:
$a + b - b = 1$
$a = 1$
So when $a + b = 1$, the possible values are $a = 1$ and $b = 0$.
For $a + b = -1$, using the equation $a - b = 1$, we get:
$a + b - (-1) = 1$
$a + b + 1 = 1$
$a + b = 0$
Using this equation and the equation $a - b = 1$, we can solve for $a$ and $b$:
From $a + b = 0$, we have $a = -b$.
Substituting this into $a - b = 1$, we get:
$-b - b = 1$
$-2b = 1$
$b = -\frac{1}{2}$
So when $a + b = -1$, the possible values are $a = -\frac{1}{2}$ and $b = -\frac{1}{2}$.
(c) Therefore, the possible values of $a$ and $b$ are $(1,0)$ and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.
To solve this problem, we will use the given equations to create a system of equations and solve it to find the values of $ab$, $a+b$, $a$, and $b$.
(a) To find all possible values of $ab$, we can use the equation $a - b = 1$ and solve for $a$ in terms of $b$. Adding $b$ to both sides of the equation gives us $a = b + 1$. Substituting this into the equation $a^3 - b^3 = 1$, we get $(b+1)^3 - b^3 = 1$. Expanding the cube on the left side, we have $b^3 + 3b^2 + 3b + 1 - b^3 = 1$. Simplifying this equation, we find $b^2 + 3b = 0$. Factoring out $b$, we have $b(b + 3) = 0$. Therefore, $b = 0$ or $b = -3$. Substituting these values back into the equation $a = b + 1$, we can find the corresponding values of $a$. So the possible values of $ab$ are $0(0 + 1) = 0$ and $-3(-3 + 1) = 6$.
(b) To find all possible values of $a + b$, we can simply add the two equations $a - b = 1$ and $a^3 - b^3 = 1$. This gives us $a + a^3 = 2$. From the first equation, we know $a = b + 1$, so substituting this gives us $b + 1 + (b + 1)^3 = 2$. Expanding the cube on the left side, we have $b + 1 + b^3 + 3b^2 + 3b + 1 = 2$. Simplifying this equation, we find $b^3 + 3b^2 + 4b = 0$. Factoring out $b$, we have $b(b^2 + 3b + 4) = 0$. However, the quadratic $b^2 + 3b + 4$ has no real solutions, so the only possible value for $b$ is $b = 0$. Substituting this back into the equation $a = b + 1$, we find $a = 1$. Therefore, the only possible value of $a + b$ is $1 + 0 = 1$.
(c) Given the values of $ab$ and $a + b$ we found in parts (a) and (b), we can determine the possible values of $a$ and $b$. From part (a), we know that $ab$ can be either $0$ or $6$. From part (b), we know that $a + b$ can be $1$.
For $ab = 0$, we can have $a = 0$ and $b = 1$, or $a = 1$ and $b = 0$.
For $ab = 6$, we can have $a = 2$ and $b = -1$, or $a = -1$ and $b = 2$.
Therefore, the possible values for $a$ and $b$ are $(a, b) = (0, 1), (1, 0), (2, -1),$ and $(-1, 2)$.