https://file.aiquickdraw.com/mathbot/file/1728940365013v6oylm8d.jpg

To solve this problem, let’s break down each function and match it with the given options.

Given:
- \( f(x) = x^3 - 27 \)
- \( g(x) = x - 3 \)

We need to evaluate each of the following expressions:

### 1. \( g(x)f(x) \)
This means multiplying \( g(x) \) and \( f(x) \):
\[
g(x)f(x) = (x - 3)(x^3 - 27)
\]
Expanding this, we get:
\[
= x \cdot x^3 - x \cdot 27 - 3 \cdot x^3 + 3 \cdot 27
\]
\[
= x^4 - 3x^3 - 27x + 81
\]
So, this matches with **Option A**: \( 81 - 27x - 3x^3 + x^4 \).

### 2. \( f(x^2) \)
This requires substituting \( x^2 \) into \( f(x) \):
\[
f(x^2) = (x^2)^3 - 27 = x^6 - 27
\]
This matches with **Option B**: \( -27 + x^6 \).

### 3. \( g(f(x)) \)
This requires substituting \( f(x) \) into \( g(x) \):
\[
g(f(x)) = f(x) - 3 = (x^3 - 27) - 3 = x^3 - 30
\]
This matches with **Option D**: \( -30 + x^3 \).

### 4. \( (g(x))^2 \)
This means squaring \( g(x) \):
\[
(g(x))^2 = (x - 3)^2 = x^2 - 6x + 9
\]
This matches with **Option C**: \( 9 - 6x + x^2 \).

### Final Answers:
1. \( g(x)f(x) \) → **A**
2. \( f(x^2) \) → **B**
3. \( g(f(x)) \) → **D**
4. \( (g(x))^2 \) → **C**

Let me know if you would like further clarification on any steps, or have questions!

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#### Related Questions:
1. How do you expand a polynomial product?
2. How do you substitute one function into another?
3. What are common methods for expanding binomials?
4. What are some tips for matching expressions with equivalent forms?
5. How does function composition differ from function multiplication?

#### Tip:
Always check for simple substitutions and expansions when working with composite functions or products.

Let f(x) = x^3 - 27 and g(x) = x - 3. Match the functions defined below with the letters labeling their equivalent expressions. 1. g(x)f(x), 2. f(x^2), 3. g(f(x)), 4. (g(x))^2.

Solve a function composition and polynomial expansion problem involving f(x) = x^3 - 27 and g(x) = x - 3. Explore detailed steps on multiplying polynomials and performing function composition. Suitable for Grades 10-12.

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