To solve the problem of expanding the expression [tex]\((a + 3b)^3\)[/tex], we will use the binomial theorem. The binomial theorem states that:
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
In this case, [tex]\(x = a\)[/tex], [tex]\(y = 3b\)[/tex], and [tex]\(n = 3\)[/tex].
Using the binomial theorem, we expand [tex]\((a + 3b)^3\)[/tex] as follows:
[tex]\[
(a + 3b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} (3b)^k
\][/tex]
Let's break down each term in this summation:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} a^{3-0} (3b)^0 = 1 \cdot a^3 \cdot 1 = a^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} a^{3-1} (3b)^1 = 3 \cdot a^2 \cdot (3b) = 3 \cdot a^2 \cdot 3b = 9a^2b
\][/tex]
3. For [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]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} a^{3-3} (3b)^3 = 1 \cdot 1 \cdot (3b)^3 = 1 \cdot 1 \cdot 27b^3 = 27b^3
\][/tex]
By summing all these terms together, we get:
[tex]\[
(a + 3b)^3 = a^3 + 9a^2b + 27ab^2 + 27b^3
\][/tex]
Therefore, the expanded form of [tex]\((a + 3b)^3\)[/tex] is:
[tex]\[
a^3 + 9a^2b + 27ab^2 + 27b^3
\][/tex]
