Sure, let's expand the expression [tex]\((2p + q)^3\)[/tex] step-by-step.
To expand [tex]\((2p + q)^3\)[/tex], we can use the binomial theorem which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, [tex]\(a = 2p\)[/tex], [tex]\(b = q\)[/tex], and [tex]\(n = 3\)[/tex].
First, identify the binomial coefficients for [tex]\(n = 3\)[/tex]:
- [tex]\(\binom{3}{0} = 1\)[/tex]
- [tex]\(\binom{3}{1} = 3\)[/tex]
- [tex]\(\binom{3}{2} = 3\)[/tex]
- [tex]\(\binom{3}{3} = 1\)[/tex]
Now, apply the binomial theorem:
[tex]\[
(2p + q)^3 = \binom{3}{0} (2p)^3 q^0 + \binom{3}{1} (2p)^2 q^1 + \binom{3}{2} (2p)^1 q^2 + \binom{3}{3} (2p)^0 q^3
\][/tex]
Simplify each term:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (2p)^{3} q^0 = 1 \cdot (2p)^3 \cdot q^0 = 8p^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} (2p)^2 q^1 = 3 \cdot (2p)^2 \cdot q = 3 \cdot 4p^2 \cdot q = 12p^2q
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (2p)^1 q^2 = 3 \cdot (2p) \cdot q^2 = 3 \cdot 2p \cdot q^2 = 6pq^2
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (2p)^0 q^3 = 1 \cdot (2p)^0 \cdot q^3 = q^3
\][/tex]
Combining all these terms, we get the expanded form of [tex]\((2p + q)^3\)[/tex]:
[tex]\[
(2p + q)^3 = 8p^3 + 12p^2q + 6pq^2 + q^3
\][/tex]
So, the expanded expression is:
[tex]\[
8p^3 + 12p^2q + 6pq^2 + q^3
\][/tex]
