To factorize the expression [tex]\(s^3 + 27t^3\)[/tex], we use the identity for the sum of cubes. Let’s go through the steps to factorize [tex]\(s^3 + 27t^3\)[/tex]:
1. Identify the sum of cubes: The given expression can be written as a sum of cubes. Notice that:
[tex]\[
s^3 + 27t^3 = s^3 + (3t)^3
\][/tex]
2. Use the sum of cubes formula: The sum of cubes [tex]\(a^3 + b^3\)[/tex] can be factored using the following formula:
[tex]\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\][/tex]
Here, [tex]\(a = s\)[/tex] and [tex]\(b = 3t\)[/tex].
3. Apply the formula: Plugging in [tex]\(s\)[/tex] and [tex]\(3t\)[/tex] into the formula, we get:
[tex]\[
s^3 + (3t)^3 = (s + 3t)((s)^2 - s(3t) + (3t)^2)
\][/tex]
4. Simplify the expression inside the parentheses:
[tex]\[
(s)^2 - s(3t) + (3t)^2 = s^2 - 3st + 9t^2
\][/tex]
5. Combine the factors: Therefore, the fully factored form of [tex]\(s^3 + 27t^3\)[/tex] is:
[tex]\[
(s + 3t)(s^2 - 3st + 9t^2)
\][/tex]
Considering the given partial factorization [tex]\(s^3 + 27t^3 = s^2 - 3st + 9t^2\)[/tex], the missing binomial that completes the factorization is:
[tex]\[
s + 3t
\][/tex]
So, the complete factorization of the given expression is:
[tex]\[
s^3 + 27t^3 = (s + 3t)(s^2 - 3st + 9t^2)
\][/tex]
