To expand the expression \((x - 3y)^3\), we can use the binomial theorem, which states:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this case, \(a = x\), \(b = -3y\), and \(n = 3\). Let's apply the binomial expansion:

\[
(x - 3y)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} (-3y)^k
\]

### Step 1: Calculate the Binomial Coefficients
\[
\binom{3}{0} = 1, \quad \binom{3}{1} = 3, \quad \binom{3}{2} = 3, \quad \binom{3}{3} = 1
\]

### Step 2: Expand the Expression
Now, we expand each term:

- When \(k = 0\):
  \[
  \binom{3}{0} x^{3-0} (-3y)^0 = 1 \cdot x^3 \cdot 1 = x^3
  \]

- When \(k = 1\):
  \[
  \binom{3}{1} x^{3-1} (-3y)^1 = 3 \cdot x^2 \cdot (-3y) = -9x^2y
  \]

- When \(k = 2\):
  \[
  \binom{3}{2} x^{3-2} (-3y)^2 = 3 \cdot x \cdot 9y^2 = 27xy^2
  \]

- When \(k = 3\):
  \[
  \binom{3}{3} x^{3-3} (-3y)^3 = 1 \cdot 1 \cdot (-27y^3) = -27y^3
  \]

### Step 3: Combine All Terms
Finally, add all the terms together:

\[
(x - 3y)^3 = x^3 - 9x^2y + 27xy^2 - 27y^3
\]

This is the expanded form of \((x - 3y)^3\).

Would you like more details or have any questions?

### Relative Questions
1. How does the binomial theorem help in expanding binomials raised to higher powers?
2. What are binomial coefficients, and how are they calculated?
3. Can you apply the binomial expansion to \((x + y)^4\)?
4. How can you use Pascal's triangle to find binomial coefficients?
5. What are the general properties of binomials raised to different powers?

### Tip
When expanding binomials using the binomial theorem, make sure to keep track of the signs carefully, especially when dealing with negative terms.

Learn how to expand the expression (x - 3y)^3 using the binomial theorem with a detailed step-by-step solution. This problem covers key concepts in algebra, including binomial coefficients and polynomial expansion. Suitable for advanced high school students in Grades 11-12.

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