Use the principle of mathematical induction to prove that $5n^3-3n^2-2n$ is divisible by 6 for all $n\in \mathbb{Z}^+$.
(1) If $n=1$, $5-3-2=0$ which is divisible by $6$. ∴$P_{1}$ is true.
(2) If $P_{k}$ is true, then $5k^3-3k^2-2k = 6A$ for some $A\in \mathbb{N}+$.
\begin{align*} 5(k+1)^3-3(k+1)^2-2(k+1) &= 5(k^3+3k^2+3k+1)-3(k^2+2k+1)-2k-2\\ &= 5k^3+15k^2+15k+5-3k^2-6k-3-2k-2\\ &= (5k^3-3k^2-2k)+15k^2+9k\\ &= 6A+15k^2+9k \end{align*}
using $P_{k}$.
I don't know where to go from here.
Continuing from what you've worked out, it is enough if we prove $15k^2 +9k$ is divisible by $6$. But $15k^2 + 9k = 3k(5k+3)$. Hence it is enough to prove $k(5k+3)$ is divisible by $2$
Now if $k$ is even, then $k$ is divisible by $2$ and we're done. If not, then $k$ is odd and $5k$ is odd being the product of two odd numbers. Then $5k+3$ will be even being the sum of two odd numbers.
Hence for every natural number $k$, $k(5k+3)$ is even and therefore $15k^2 + 9k = 3k(5k+3)$ is divisible by $6$, which completes the proof. $\square$
Use the principle of mathematical induction to prove that $5n^3-3n^2-2n$ is divisible by 6 for all $n\in \mathbb{Z}^+$.
Alternative approach:
Forgo elegance, manually check whether the assertion is true for each element in $\{1,2,3,4,5,6\}$, and then prove that if the assertion is true for $n$, then the assertion is also true for $(n+6).$ This will establish that the assertion is true for each element in $\{7,\cdots,12\}$, which will establish that the assertion is true for each element in $\{13,\cdots,18\}$, and so forth.
Let $f(n) = 5n^3 - 3n^2 - 2n \implies$
$f(1) = 0, ~f(2) = 24, ~f(3) = 102,$
$f(4) = 264, ~f(5) = 540.$
Further, clearly $~6 ~| ~f(6),~$ since each of the three terms in $f(6)$ is a multiple of $~6$.
Therefore, the the assertion is true for each element in $\{1,2,3,4,5,6\}.$
The remainder of the problem is resolved by the general theorem, that for any positive integers $n,k,r$ you have that $~k~$ divides $~(n+k)^r - n^r.$
This result is immediate by the binomial theorem, since every term in the summation $\displaystyle \left[ ~\sum_{i=1}^r \binom{r}{i}k^i n^{r-i} ~\right]~$ is divisible by $k$.
This establishes that (for example) $6$ divides each of
$\left[ ~(n+6)^3 - n^3 ~\right], ~\left[ ~(n+6)^2 - n^2 ~\right], ~\left[ ~(n+6)^1 - n^1 ~\right].$
Therefore, for each positive integer $n$:
If $6 ~| ~f(n) ~$ then $6 ~| ~f(n+6).$
