To prove that an odd number cubed is still an odd number, let's express an odd number in its general form and then cube it.
An odd number can be expressed in the form [tex]\(2n + 1\)[/tex], where [tex]\(n\)[/tex] is any integer. Now, let's cube this number:
[tex]\[
(2n + 1)^3
\][/tex]
Next, we need to expand this expression:
[tex]\[
(2n + 1)^3 = (2n + 1)(2n + 1)(2n + 1)
\][/tex]
Let's first find [tex]\((2n + 1)^2\)[/tex]:
[tex]\[
(2n + 1)^2 = (2n + 1)(2n + 1) = 4n^2 + 4n + 1
\][/tex]
Now, we multiply this result by [tex]\((2n + 1)\)[/tex]:
[tex]\[
(2n + 1)(4n^2 + 4n + 1)
\][/tex]
We distribute [tex]\((2n + 1)\)[/tex] across each term inside the parenthesis:
[tex]\[
= (2n + 1)(4n^2) + (2n + 1)(4n) + (2n + 1)(1)
\][/tex]
[tex]\[
= 2n \cdot 4n^2 + 1 \cdot 4n^2 + 2n \cdot 4n + 1 \cdot 4n + 2n \cdot 1 + 1 \cdot 1
\][/tex]
[tex]\[
= 8n^3 + 4n^2 + 8n^2 + 4n + 2n + 1
\][/tex]
Combining like terms:
[tex]\[
= 8n^3 + 12n^2 + 6n + 1
\][/tex]
So,
[tex]\[
(2n + 1)^3 = 8n^3 + 12n^2 + 6n + 1
\][/tex]
Notice that [tex]\(8n^3\)[/tex], [tex]\(12n^2\)[/tex], and [tex]\(6n\)[/tex] are all even numbers because they are multiples of 2.
Adding these even numbers [tex]\(8n^3 + 12n^2 + 6n\)[/tex] together results in an even number. We then add 1, which is an odd number, to this even sum:
[tex]\[
8n^3 + 12n^2 + 6n + 1
\][/tex]
The final expression is an even number plus 1, which is an odd number. Therefore, we have shown that the cube of an odd number [tex]\((2n + 1)^3\)[/tex] is still an odd number.
