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The image shows the expression \((x - y)^5\). To expand this expression, we can apply the **Binomial Theorem**, which states that:

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

In this case, we treat \(a = x\), \(b = -y\), and \(n = 5\). Therefore, we expand \((x - y)^5\) as follows:

\[
(x - y)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-y)^k
\]

Breaking this down step by step:

1. For \(k = 0\):
   \[
   \binom{5}{0} x^5 (-y)^0 = 1 \cdot x^5 \cdot 1 = x^5
   \]

2. For \(k = 1\):
   \[
   \binom{5}{1} x^4 (-y)^1 = 5 \cdot x^4 \cdot (-y) = -5x^4y
   \]

3. For \(k = 2\):
   \[
   \binom{5}{2} x^3 (-y)^2 = 10 \cdot x^3 \cdot y^2 = 10x^3y^2
   \]

4. For \(k = 3\):
   \[
   \binom{5}{3} x^2 (-y)^3 = 10 \cdot x^2 \cdot (-y)^3 = -10x^2y^3
   \]

5. For \(k = 4\):
   \[
   \binom{5}{4} x^1 (-y)^4 = 5 \cdot x \cdot y^4 = 5xy^4
   \]

6. For \(k = 5\):
   \[
   \binom{5}{5} x^0 (-y)^5 = 1 \cdot 1 \cdot (-y)^5 = -y^5
   \]

Putting all the terms together:

\[
(x - y)^5 = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5
\]

This is the expanded form of the given expression.

Would you like more details on any part of this process? Here are five related questions that can help you understand more:

1. How does the Binomial Theorem apply to powers higher than 5?
2. How are binomial coefficients calculated for different powers?
3. How does the sign change affect the expansion when dealing with \((x - y)^n\)?
4. What is Pascal's Triangle, and how does it relate to binomial expansion?
5. Can we apply a similar expansion to fractional or negative exponents?

**Tip**: Binomial expansion is particularly useful for approximating expressions in calculus, especially when using series expansions like Taylor and Maclaurin series!

Explore the binomial expansion of (x - y)^5, applying the Binomial Theorem to expand the expression into its polynomial form. This problem introduces key concepts in algebra, such as binomial coefficients and polynomial expansion, suitable for students in Grades 10-12.

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