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We are given the profit function \(\pi(x)\) of a company as follows:

\[
\pi(x) = (x + 5)^3 - (x - k)^2
\]

We are also told that \(x = -1\) is a root of the profit function, which means that:

\[
\pi(-1) = 0
\]

We need to find the product of the values of \(k\).

### Step 1: Substitute \(x = -1\) into the profit function.
\[
\pi(-1) = (-1 + 5)^3 - (-1 - k)^2
\]
Simplifying both terms:

\[
(-1 + 5)^3 = 4^3 = 64
\]
\[
(-1 - k)^2 = (-(1 + k))^2 = (1 + k)^2
\]

Thus, the equation becomes:

\[
64 - (1 + k)^2 = 0
\]

### Step 2: Solve for \(k\).
Rearrange the equation to isolate \((1 + k)^2\):

\[
64 = (1 + k)^2
\]

Take the square root of both sides:

\[
1 + k = \pm 8
\]

### Step 3: Solve for \(k\) in both cases.
1. \(1 + k = 8\) gives \(k = 7\).
2. \(1 + k = -8\) gives \(k = -9\).

### Step 4: Find the product of the values of \(k\).
The values of \(k\) are \(7\) and \(-9\). The product is:

\[
7 \times (-9) = -63
\]

Thus, the correct answer is \( \boxed{-63} \).

---

Would you like a more detailed explanation, or do you have any questions?

Here are five related questions to further your understanding:
1. How do we identify a root of a function and use it to solve for unknowns?
2. Why do we take both the positive and negative square roots in Step 2?
3. Can you solve for \(k\) if the equation was \(64 + (1 + k)^2 = 0\)?
4. How would this problem change if the profit function had a different exponent, e.g., \((x - k)^3\)?
5. What are other scenarios where solving for \(k\) would involve factoring instead of square roots?

**Tip:** Always check if multiple solutions are possible when dealing with quadratic equations!

The profit (π) of a company is determined by the excess of income over costs: π(x) = (x + 5)^3 - (x - k)^2. If -1 is a root of the expression, find the product of the values of k.

Solve for the values of k in the profit equation π(x) = (x + 5)^3 - (x - k)^2, knowing that -1 is a root. Learn how to apply algebraic methods, solve quadratic equations, and find the product of the solutions.

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