To expand the expression [tex]\((3x + 4y)^3\)[/tex], we use the binomial theorem. The binomial theorem states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Applying this theorem to our expression [tex]\((3x + 4y)^3\)[/tex], we have:
- [tex]\((3x)\)[/tex] as [tex]\(a\)[/tex]
- [tex]\((4y)\)[/tex] as [tex]\(b\)[/tex]
- [tex]\(3\)[/tex] as [tex]\(n\)[/tex]
Let’s expand [tex]\((3x + 4y)^3\)[/tex] step-by-step using the binomial theorem.
1. Identify the terms and coefficients using the binomial theorem:
[tex]\[
(3x + 4y)^3 = \sum_{k=0}^{3} \binom{3}{k} (3x)^{3-k} (4y)^k
\][/tex]
Here, [tex]\(\binom{n}{k}\)[/tex] is a binomial coefficient which can be calculated as [tex]\(\binom{3}{k} = \frac{3!}{k!(3-k)!}\)[/tex].
2. Calculate each term in the sum:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (3x)^{3-0} (4y)^0 = 1 \cdot (3x)^3 \cdot 1 = 27x^3
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} (3x)^{3-1} (4y)^1 = 3 \cdot (3x)^2 \cdot (4y) = 3 \cdot 9x^2 \cdot 4y = 108x^2y
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (3x)^{3-2} (4y)^2 = 3 \cdot (3x)^1 \cdot (4y)^2 = 3 \cdot 3x \cdot 16y^2 = 144xy^2
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (3x)^{3-3} (4y)^3 = 1 \cdot (3x)^0 \cdot (4y)^3 = 1 \cdot 1 \cdot 64y^3 = 64y^3
\][/tex]
3. Combine all the terms:
Adding all the computed terms together:
[tex]\[
(3x + 4y)^3 = 27x^3 + 108x^2y + 144xy^2 + 64y^3
\][/tex]
So, the expanded form of [tex]\((3x + 4y)^3\)[/tex] is:
[tex]\[
27x^3 + 108x^2y + 144xy^2 + 64y^3
\][/tex]
