Question

prove: an odd integer cubed is odd. (2n + 1)³ = ?n³ + □n² + □n + □ = □(4n³ + 6n² + 3n) + 1 = odd

Explanation:

Step1: Expand \((2n + 1)^3\) using \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

Here, \(a = 2n\) and \(b = 1\), so:
\[

$$\begin{align*} (2n + 1)^3&=(2n)^3+3\times(2n)^2\times1 + 3\times(2n)\times1^2+1^3\\ &=8n^3 + 12n^2+6n + 1 \end{align*}$$

\]

Step2: Factor the first three terms

Notice that \(8n^3 + 12n^2+6n = 2\times(4n^3 + 6n^2 + 3n)\), so:
\[
(2n + 1)^3=2\times(4n^3 + 6n^2 + 3n)+1
\]

Answer:

The first blank (coefficient of \(n^3\)) is \(8\), the second (coefficient of \(n^2\)) is \(12\), the third (coefficient of \(n\)) is \(6\), the fourth (constant term) is \(1\), and the blank in the factoring step is \(2\). So filling in the blanks: \((2n + 1)^3=\boldsymbol{8}n^3+\boldsymbol{12}n^2+\boldsymbol{6}n+\boldsymbol{1}=\boldsymbol{2}(4n^3 + 6n^2 + 3n)+1\)