$n^3 - (n-1)^3$

$\begin{aligned} n^3-(n-1)^3 & = n^3 - (n^3-3n^2+3n-1) \\ & = 3n^2 - 3n + 1 \\ & = \underbrace{3\overbrace{n(n-1)}^{\text{Always even}} + 1}_{\text{Always odd}} \end{aligned}$

Note that $n(n-1)$ is always even, because either $n$ or $n-1$ is even. Therefore $3n(n+1) + 1$ is always odd. True

$n^3-(n-1)^3 \equiv n-(n-1) = 1 \pmod{2}$

Write out $n^3 - (n-1)^3$ in full

= $n^3 - n^3 + 3n^2 - 3n + 1$

= $3n^2 - 3n + 1$

If $n$ is even, let $n = 2c$ , where $c$ is an integer, and replace all $n$ 's with $2c$

Odd numbers can be expressed as $2c + 1$

= $3(2c)^2$ $-$ $3(2c)$ $+ 1$

= $3(4c^2) - 6c + 1$

= $12c^2 - 6c + 1$

= $2(6c^2 - 3c) + 1$ This is in the form of an odd number, so if $n$ is even, $n^3 - (n-1)^3$ is odd.

If $n$ is odd, let $n = 2c+1$ , where c is an integer.

= $3(2c+1)^2 - 3(2c+1) + 1$

= $12c^2 + 1$

= $2(6c^2) + 1$ Again, this is in the form of an odd number, so if $n$ is odd, $n^3 - (n-1)^3$ is odd.

Since $n^3 - (n-1)^3$ is odd when $n$ is odd & $n$ is even, $n^3 - (n-1)^3$ is always odd when $n$ is an integer.