To solve the expression [tex]\((3c + 4m)^3\)[/tex], we'll expand it using the binomial theorem. According to the binomial theorem, [tex]\((a + b)^n\)[/tex] can be expanded as:
[tex]\[
(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\(a = 3c\)[/tex], [tex]\(b = 4m\)[/tex], and [tex]\(n = 3\)[/tex]. Let's use the binomial theorem to expand [tex]\((3c + 4m)^3\)[/tex].
1. First Term: When [tex]\(k = 0\)[/tex]
[tex]\[
\binom{3}{0} (3c)^{3-0} (4m)^0 = 1 \cdot (3c)^3 \cdot 1 = 27c^3
\][/tex]
2. Second Term: When [tex]\(k = 1\)[/tex]
[tex]\[
\binom{3}{1} (3c)^{3-1} (4m)^1 = 3 \cdot (3c)^2 \cdot 4m = 3 \cdot 9c^2 \cdot 4m = 108c^2m
\][/tex]
3. Third Term: When [tex]\(k = 2\)[/tex]
[tex]\[
\binom{3}{2} (3c)^{3-2} (4m)^2 = 3 \cdot (3c)^1 \cdot (4m)^2 = 3 \cdot 3c \cdot 16m^2 = 144cm^2
\][/tex]
4. Fourth Term: When [tex]\(k = 3\)[/tex]
[tex]\[
\binom{3}{3} (3c)^{3-3} (4m)^3 = 1 \cdot 1 \cdot (4m)^3 = 64m^3
\][/tex]
Now, combine all these terms together:
[tex]\[
(3c + 4m)^3 = 27c^3 + 108c^2m + 144cm^2 + 64m^3
\][/tex]
Thus, the expanded form of [tex]\((3c + 4m)^3\)[/tex] is:
[tex]\[
27c^3 + 108c^2m + 144cm^2 + 64m^3
\][/tex]
