To solve the expression [tex]\((3m - 2n)^3\)[/tex], we need to expand it using the binomial theorem for expanding powers of binomials. The binomial theorem states that:
[tex]\[
(a + b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} b^{k}
\][/tex]
In our case, [tex]\( a = 3m \)[/tex] and [tex]\( b = -2n \)[/tex]. Let's expand step-by-step:
Step 1: Identify the binomial coefficients for powers of 3:
[tex]\[
\binom{3}{0} = 1, \quad \binom{3}{1} = 3, \quad \binom{3}{2} = 3, \quad \binom{3}{3} = 1
\][/tex]
Step 2: Write out each term of the expansion:
[tex]\[
(3m - 2n)^3 = \binom{3}{0} (3m)^3 (-2n)^0 + \binom{3}{1} (3m)^2 (-2n)^1 + \binom{3}{2} (3m)^1 (-2n)^2 + \binom{3}{3} (3m)^0 (-2n)^3
\][/tex]
Step 3: Compute each term separately:
1. For [tex]\(\binom{3}{0} (3m)^3 (-2n)^0\)[/tex]:
[tex]\[
\binom{3}{0} = 1, \quad (3m)^3 = 27m^3, \quad (-2n)^0 = 1 \quad \Rightarrow \quad 1 \cdot 27m^3 \cdot 1 = 27m^3
\][/tex]
2. For [tex]\(\binom{3}{1} (3m)^2 (-2n)^1\)[/tex]:
[tex]\[
\binom{3}{1} = 3, \quad (3m)^2 = 9m^2, \quad (-2n)^1 = -2n \quad \Rightarrow \quad 3 \cdot 9m^2 \cdot (-2n) = -54m^2n
\][/tex]
3. For [tex]\(\binom{3}{2} (3m)^1 (-2n)^2\)[/tex]:
[tex]\[
\binom{3}{2} = 3, \quad (3m)^1 = 3m, \quad (-2n)^2 = 4n^2 \quad \Rightarrow \quad 3 \cdot 3m \cdot 4n^2 = 36mn^2
\][/tex]
4. For [tex]\(\binom{3}{3} (3m)^0 (-2n)^3\)[/tex]:
[tex]\[
\binom{3}{3} = 1, \quad (3m)^0 = 1, \quad (-2n)^3 = -8n^3 \quad \Rightarrow \quad 1 \cdot 1 \cdot (-8n^3) = -8n^3
\][/tex]
Step 4: Combine all the terms:
[tex]\[
(3m - 2n)^3 = 27m^3 - 54m^2n + 36mn^2 - 8n^3
\][/tex]
So, the expanded form of [tex]\((3m - 2n)^3\)[/tex] is:
[tex]\[
(3m - 2n)^3 = 27m^3 - 54m^2n + 36mn^2 - 8n^3
\][/tex]
