To expand the expression [tex]\((a + b)^3\)[/tex], we apply the binomial theorem, which states that:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
For [tex]\( n = 3 \)[/tex], the expansion is:
[tex]\[ (a + b)^3 = \binom{3}{0} a^3 b^0 + \binom{3}{1} a^2 b + \binom{3}{2} a b^2 + \binom{3}{3} a^0 b^3 \][/tex]
Let's compute each term individually:
- [tex]\(\binom{3}{0}\)[/tex] is the binomial coefficient "3 choose 0," which equals 1. Thus, the first term is [tex]\(1 \cdot a^3 \cdot b^0 = a^3\)[/tex].
- [tex]\(\binom{3}{1}\)[/tex] is the binomial coefficient "3 choose 1," which equals 3. Thus, the second term is [tex]\(3 \cdot a^2 \cdot b = 3a^2b\)[/tex].
- [tex]\(\binom{3}{2}\)[/tex] is the binomial coefficient "3 choose 2," which also equals 3. Thus, the third term is [tex]\(3 \cdot a \cdot b^2 = 3ab^2\)[/tex].
- [tex]\(\binom{3}{3}\)[/tex] is the binomial coefficient "3 choose 3," which equals 1. Thus, the fourth term is [tex]\(1 \cdot a^0 \cdot b^3 = b^3\)[/tex].
Putting these terms together, we get the expanded form of the expression:
[tex]\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \][/tex]
