We use two tricks to prove this statement. One, is that if two numbers are multiples of 9, then so are their differences. Second, a difference of cubes has a nice formula that we utilize with the 1st trick: \(a^3-b^3=(a-b)(a^2+ab+b^2)\):
$$
\text{Proof that: }f(k)=k^3+(k+1)^3+(k+2)^3=9n\text{, where }k,n\in\Bbb N\\
\text{Basis:}\\
\quad f(0)=1^3+2^3=9\text{, which is a multiple of 9 with }n=1\\\\
\text{Induction Step:}\\\\
\quad\text{Assume }f(k)=9m\text{,where }k,m\in\Bbb N\\
\quad f(k+1)=9p\equiv f(k+1)-f(k)=9q\text{, where }k,p,q\in\Bbb N\\
\quad f(k+1)=(k+1)^3+(k+2)^3+(k+3)^3\\
\quad\text{So, }f(k+1)-f(k)\\
\quad=(k+3)^3-k^3\\
\quad=(k+3-k)((k+3)^2+k(k+3)+k^2)\\
\quad=3(k^2+6k+9+k^2+3k+k^2)\\
\quad=3(3k^2+9k+9)\\
\quad=3(3)(k^2+3k+1)\\
\quad=9q\text{, where }q\in\Bbb N\\\\
\text{Therefore, }f(k+1)=9p\text{, QED.}
$$