To find the expansion of [tex]\((s + t)^3\)[/tex], we will use the binomial theorem, which 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 a binomial coefficient defined as [tex]\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)[/tex].
Here, [tex]\(a = s\)[/tex], [tex]\(b = t\)[/tex], and [tex]\(n = 3\)[/tex]. Plugging in these values, we have:
[tex]$(s + t)^3 = \sum_{k=0}^{3} \binom{3}{k} s^{3-k} t^k$[/tex]
Now, we will expand this sum term by term:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} s^{3-0} t^0 = \binom{3}{0} s^3 t^0 = 1 \cdot s^3 \cdot 1 = s^3
\][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} s^{3-1} t^1 = \binom{3}{1} s^2 t = 3 \cdot s^2 \cdot t = 3s^2 t
\][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} s^{3-2} t^2 = \binom{3}{2} s t^2 = 3 \cdot s \cdot t^2 = 3s t^2
\][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} s^{3-3} t^3 = \binom{3}{3} t^3 = 1 \cdot 1 \cdot t^3 = t^3
\][/tex]
Combining all these terms, we get:
[tex]\[
(s + t)^3 = s^3 + 3s^2 t + 3s t^2 + t^3
\][/tex]
Thus, the expanded form of [tex]\((s + t)^3\)[/tex] is:
[tex]\[
(s + t)^3 = s^3 + 3s^2t + 3st^2 + t^3
\][/tex]
