Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!

Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$

Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^3+(k+1)^3+(k+2)^3+(k+3)^3 = 4b$ where $b$ is some integer. We need to show $4$ divides $(k+1)^3+(k+2)^3+(k+3)^3+(k+4)^3$.
\begin{align}
(k+1)^3 & +(k+2)^3+(k+3)^3+(k+4)^3\\ 
&= 4k^3+30k^2+90k+100\\
&=(k^3+(k+1)^3+(k+2)^3+(k+3)^3)+12k^2+48k+36\\
&=4b+12k^2+48k+36 \qquad \text{(by inductive hypothesis)}\\
&=4(b+3k^2+12k+9)
\end{align}
Since $b$ is an element of any integer this holds true for $(k+1)$. Hence proven.

Your answer seems right and looks like a short route, I would use modules to avoid the algebra but it wouldn't be induction.
For $K \equiv 0 \pmod 4$ the modules of $4$ for the elements of the equation would be $0^3 + 1^3 + 2^3 + (-1)^3 \equiv 0 + 1 + 0 + (-1) = 0 \pmod 4$.
And because there are always $4$ consecutive elements, the four module values are going to be the same for any $K$.
