To find the expanded form of [tex]\((7x + 5)^3\)[/tex], we need to employ 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], where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient. In this particular case, [tex]\(a = 7x\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(n = 3\)[/tex].
Using the binomial theorem, we can expand [tex]\((7x + 5)^3\)[/tex] as follows:
[tex]\[ (7x + 5)^3 = \sum_{k=0}^{3} \binom{3}{k} (7x)^{3-k} 5^k \][/tex]
Let's break this down term by term:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} (7x)^3 5^0 = 1 \cdot (7x)^3 \cdot 1 = (7x)^3 = 343x^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} (7x)^2 5^1 = 3 \cdot (7x)^2 \cdot 5 = 3 \cdot 49x^2 \cdot 5 = 735x^2
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} (7x)^1 5^2 = 3 \cdot 7x \cdot 25 = 3 \cdot 7x \cdot 25 = 525x
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} (7x)^0 5^3 = 1 \cdot 1 \cdot 125 = 125
\][/tex]
Now, combining all these terms together, we get:
[tex]\[ (7x + 5)^3 = 343x^3 + 735x^2 + 525x + 125 \][/tex]
Thus, the fully expanded form of [tex]\((7x + 5)^3\)[/tex] is:
[tex]\[ 343x^3 + 735x^2 + 525x + 125 \][/tex]
