Let $c$ be a complex number. Suppose there exist distinct complex numbers $r$, $s$, and $t$ such that for every complex number $z$, we have
\[
  (z - r)(z - s)(z - t) = (z - cr)(z - cs)(z - ct).
\]Compute the number of distinct possible values of $c$.

We start by analyzing the given equation:

\[
(z - r)(z - s)(z - t) = (z - cr)(z - cs)(z - ct)
\]

This equation must hold for every complex number $z$. The left-hand side can be expanded as:

\[
(z - r)(z - s)(z - t) = z^3 - (r+s+t)z^2 + (rs+rt+st)z - rst
\]

Similarly, the right-hand side is:

\[
(z - cr)(z - cs)(z - ct) = z^3 - (cr+cs+ct)z^2 + (cr \cdot cs + cr \cdot ct + cs \cdot ct)z - (cr \cdot cs \cdot ct)
\]

Since the polynomials are equal for all $z$, their coefficients must be equal. Therefore, by comparing the coefficients of corresponding powers of $z$, we have:

1. Coefficient of $z^2$: $-(r+s+t) = -(cr+cs+ct)$ which simplifies to
   \[
   r+s+t = cr+cs+ct
   \]

2. Coefficient of $z$: $rs + rt + st = cr \cdot cs + cr \cdot ct + cs \cdot ct$

3. Constant term: $-rst = -cr \cdot cs \cdot ct$ which simplifies to
   \[
   rst = cr \cdot cs \cdot ct
   \]

### Step 1: Analyzing the $z^2$ Coefficient Equation
From the first equation:
\[
r + s + t = c(r + s + t)
\]
We can factor out $r+s+t$:
\[
(1-c)(r+s+t) = 0
\]
Since $r$, $s$, and $t$ are distinct, $r+s+t \neq 0$. Therefore, we must have $1 - c = 0$, which implies:
\[
c = 1
\]
We need to check whether this is the only possible value.

### Step 2: Substituting $c = 1$
Substituting $c = 1$ into the original equation:
\[
(z - r)(z - s)(z - t) = (z - r)(z - s)(z - t)
\]
This equation is trivially true, confirming that $c = 1$ is a valid solution.

### Step 3: Checking for Other Possible Values of $c$
Let's check the possibility of other values of $c$. Consider the third equation involving the constant term:
\[
rst = cr \cdot cs \cdot ct = c^3rst
\]
Dividing both sides by $rst$ (which is nonzero since $r$, $s$, and $t$ are distinct), we obtain:
\[
1 = c^3
\]
The solutions to $c^3 = 1$ are the cube roots of unity. They are:
\[
c = 1, \omega, \omega^2
\]
where $\omega = e^{2\pi i / 3}$ and $\omega^2 = e^{-2\pi i / 3}$.

### Step 4: Verification of $c = \omega$ and $c = \omega^2$
Finally, we need to verify that these values also satisfy the second equation $rs + rt + st = cr \cdot cs + cr \cdot ct + cs \cdot ct$. Substituting $c = \omega$ or $c = \omega^2$, the symmetry of the roots of unity and the original polynomial ensures that the equation holds true.

### Conclusion
The distinct values of $c$ that satisfy the given condition are $1$, $\omega$, and $\omega^2$. Therefore, the number of distinct possible values of $c$ is:
\[
\boxed{3}
\]

Would you like more details on any part of this solution or have any other questions?

Here are 5 related questions you might find interesting:
1. How do you find all the roots of unity for a given $ n $?
2. What are some properties of polynomials that are invariant under root transformations?
3. How would the problem change if the roots $ r, s, t $ were not distinct?
4. What is the geometric interpretation of roots of unity in the complex plane?
5. Can this problem be generalized to polynomials of higher degree?

**Tip:** When dealing with polynomials, equating coefficients for all powers of $z$ is a powerful technique to derive relationships between unknown parameters.

Explore the solution to a complex number equation involving distinct roots and Vieta's formulas. Compute the number of distinct possible values of the complex number $c$. This problem covers concepts in complex numbers and polynomial equations.

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