To solve the expression [tex]\((a + 3b)^3\)[/tex], we need to expand it completely. Here is the step-by-step process:
1. Use the binomial theorem:
The binomial theorem states that [tex]\((x + y)^n\)[/tex] can be expanded as the sum of the terms of the form [tex]\(\binom{n}{k} x^{n-k} y^k\)[/tex], where [tex]\(\binom{n}{k}\)[/tex] is a binomial coefficient and [tex]\(k\)[/tex] ranges from 0 to [tex]\(n\)[/tex].
2. Apply the binomial theorem for [tex]\( (a + 3b)^3 \)[/tex]:
In our case, [tex]\(x = a\)[/tex], [tex]\(y = 3b\)[/tex], and [tex]\(n = 3\)[/tex].
[tex]\[(a + 3b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} (3b)^k\][/tex]
3. Calculate each term individually:
- When [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} a^{3-0} (3b)^0 = 1 \cdot a^3 \cdot 1 = a^3
\][/tex]
- When [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} a^{3-1} (3b)^1 = 3 \cdot a^2 \cdot 3b = 9a^2b
\][/tex]
- When [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} a^{3-2} (3b)^2 = 3 \cdot a \cdot (3b)^2 = 3 \cdot a \cdot 9b^2 = 27ab^2
\][/tex]
- When [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} a^{3-3} (3b)^3 = 1 \cdot 1 \cdot (3b)^3 = (3b)^3 = 27b^3
\][/tex]
4. Combine all the terms:
[tex]\[
(a + 3b)^3 = a^3 + 9a^2b + 27ab^2 + 27b^3
\][/tex]
So the expanded form of [tex]\((a + 3b)^3\)[/tex] is:
[tex]\[
a^3 + 9a^2b + 27ab^2 + 27b^3
\][/tex]
