Sure! Let's expand the expression [tex]\((a + 2b)^3\)[/tex] step by step using 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], where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient.
For our expression [tex]\((a + 2b)^3\)[/tex], we can set it up as follows:
[tex]\[ (a + 2b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} (2b)^k \][/tex]
Let's compute each term in the expansion step by step.
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{3}{0} a^{3-0} (2b)^0 = 1 \cdot a^3 \cdot 1 = a^3 \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{3}{1} a^{3-1} (2b)^1 = 3 \cdot a^2 \cdot (2b) = 6a^2b \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{3}{2} a^{3-2} (2b)^2 = 3 \cdot a \cdot (2b)^2 = 3 \cdot a \cdot 4b^2 = 12ab^2 \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{3}{3} a^{3-3} (2b)^3 = 1 \cdot 1 \cdot (2b)^3 = 1 \cdot 8b^3 = 8b^3 \][/tex]
Now, compile all these terms together:
[tex]\[ (a + 2b)^3 = a^3 + 6a^2b + 12ab^2 + 8b^3 \][/tex]
Therefore, [tex]\((a+2b)^3 = a^3 + 6a^2b + 12ab^2 + 8b^3\)[/tex].
