To expand [tex]\((2x + y)^3\)[/tex], we will use the binomial theorem. The binomial theorem states that [tex]\((a + b)^n\)[/tex] can be expanded into the sum of terms of the form [tex]\(\binom{n}{k} a^{n-k} b^k\)[/tex], where [tex]\(\binom{n}{k}\)[/tex] is a binomial coefficient.
We start with [tex]\((2x + y)^3\)[/tex]. Here, [tex]\(a = 2x\)[/tex], [tex]\(b = y\)[/tex], and [tex]\(n = 3\)[/tex]. Applying the binomial theorem, we get:
[tex]\[
(2x + y)^3 = \sum_{k=0}^{3} \binom{3}{k} (2x)^{3-k} y^k
\][/tex]
We can break this down term by term:
1. When [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (2x)^{3-0} y^0 = 1 \cdot (2x)^3 \cdot 1 = 8x^3
\][/tex]
2. When [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]
3. When [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (2x)^{3-2} y^2 = 3 \cdot (2x) \cdot y^2 = 3 \cdot 2x \cdot y^2 = 6xy^2
\][/tex]
4. When [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (2x)^{3-3} y^3 = 1 \cdot (2x)^0 \cdot y^3 = 1 \cdot 1 \cdot y^3 = y^3
\][/tex]
Combining all these terms, we obtain:
[tex]\[
(2x + y)^3 = 8x^3 + 12x^2y + 6xy^2 + y^3
\][/tex]
So, the expanded form of [tex]\((2x + y)^3\)[/tex] is:
[tex]\[
8x^3 + 12x^2y + 6xy^2 + y^3
\][/tex]
