To solve for [tex]\((2x - y)^3\)[/tex], we need to expand the expression using the binomial theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, we have [tex]\(a = 2x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 3\)[/tex].
Let's expand [tex]\((2x - y)^3\)[/tex] step-by-step:
1. Identify our terms:
- [tex]\(a = 2x\)[/tex]
- [tex]\(b = -y\)[/tex]
- [tex]\(n = 3\)[/tex]
2. Apply the binomial theorem:
[tex]\[
(2x - y)^3 = \sum_{k=0}^{3} \binom{3}{k} (2x)^{3-k} (-y)^k
\][/tex]
Now, let's calculate each term in the sum separately.
3. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (2x)^{3-0} (-y)^0 = 1 \cdot (2x)^3 \cdot 1 = 8x^3
\][/tex]
4. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} (2x)^{3-1} (-y)^1 = 3 \cdot (2x)^2 \cdot (-y) = 3 \cdot 4x^2 \cdot (-y) = -12x^2y
\][/tex]
5. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (2x)^{3-2} (-y)^2 = 3 \cdot (2x)^1 \cdot y^2 = 3 \cdot 2x \cdot y^2 = 6xy^2
\][/tex]
6. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (2x)^{3-3} (-y)^3 = 1 \cdot (2x)^0 \cdot (-y)^3 = -y^3
\][/tex]
7. Finally, sum all these terms together:
[tex]\[
(2x - y)^3 = 8x^3 - 12x^2y + 6xy^2 - y^3
\][/tex]
So, the fully expanded form of [tex]\((2x - y)^3\)[/tex] is:
[tex]\[
8x^3 - 12x^2y + 6xy^2 - y^3
\][/tex]
