To expand the expression [tex]\((x - y)^3\)[/tex], we can apply the binomial theorem, which states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, the expression is [tex]\((x - y)^3\)[/tex]. Here, [tex]\(a = x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 3\)[/tex].
Applying the binomial theorem, we have:
[tex]\[
(x + (-y))^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} (-y)^k
\][/tex]
Now we will expand this sum step-by-step for each value of [tex]\(k\)[/tex] from 0 to 3:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} x^{3-0} (-y)^0 = 1 \cdot x^3 \cdot 1 = x^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} x^{3-1} (-y)^1 = 3 \cdot x^2 \cdot (-y) = -3x^2y
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} x^{3-2} (-y)^2 = 3 \cdot x \cdot y^2 = 3xy^2
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} x^{3-3} (-y)^3 = 1 \cdot 1 \cdot (-y)^3 = -y^3
\][/tex]
Combining all these terms, we get the expanded form of the original expression:
[tex]\[
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
\][/tex]
So the fully expanded form of [tex]\((x - y)^3\)[/tex] is:
[tex]\[
x^3 - 3x^2y + 3xy^2 - y^3
\][/tex]
