Complete induction:figure out the last step I tried to solve this task on complete induction. I am sorry for the form, if you see a   $6$
it means: a divisible through six. Could anyone please check if I have done right? Because at the end, I don´t know how to show that  4n³+2n²+3n is divisible through six.
Please show by complete induction, that  for every $n\in\mathbb N$:  $(n²)² -n³ -n²+n$ is divisible through $6$
1) $\frac{(1²)²-1³ -1²+1 }{6}=\frac06= 0$   true statement
2) inductive hypothesis:
$\frac{(n²)²-n³-n²+n }{6}$
3) inductive step:
$n\gt n+1$
$$\frac{((n+1)²)²-(n+1)³-(n+1)²+(n+1)}{6}=$$ 
$$\frac{(n+1)(n+1)(n+1)(n+1)-(n+1)(n+1)(n+1)-(n+1)(n+1)+(n+1)}{6}=$$   
$$\frac{(n²+2n+1)(n²+2n+1)–(n²+2n+1)(n+1)-(n²+2n+1)+(n+1)}{6}=$$  
$$\frac{(n²)²+ 2n³+n² +2n³+4n²+2n+1–(n²+2n+1)(n+1)-(n²+2n+1)+(n+1)}{6}=$$   
$$\frac{(n²)²+4n³+5n²+2n+1-(n³+n²+2n²+2n+n+1)-(n²+2n+1)+(n+1)}{ 6}=$$
$$4(n²)²-n³-n²-2n$$
=inductive
hypothesis
$+ 4n³+2n²+3n $ ??? (divisible through six?)
 A: Hint $\rm\ f(n)\, =\, (n\!-\!1)^2n(n\!+\!1)\:$ has as factor a product of $3$ consecutive integers, one of which is divisible by $2$, and one divisible by $3\,$ (this can be proved by induction if that way is required).
Alternatively, $\rm\displaystyle\ \ \frac{f(n)}6\, =\, \frac{(n\!-\!1)^2n(n\!+\!1)}6\, =\, (n\!-\!1){n\!+\!1\choose 3}\,\in\,\Bbb Z$
Or, $\rm\,\ f(n) = (n\!-\!1)(n^3\!-\!n),\:$ and $\rm\ n^3\!\equiv n\,\ mod\,\ 3,\,2\ $ by little Fermat.
Remark $\ $ If you fix the errors in your calculations you'll end trying to prove as your inductive step that $\rm\ 6\mid f(n\!+\!1)\!-\!f(n) = 4n^3\!+3n^2\!-n = (4n\!-\!1)n(n\!+\!1) = (n\!-\!1)n(n\!+\!1) + 3n^2(n\!+\!1).\:$ This can be done as above, since $\,2\,$ divides one of $\rm\,n,\,n\!+\!1,\:$ and $\rm\,3\,$ divides one of $\rm\:n\!-\!1,\,n,\,n\!+\!1.\:$ 
Calculating the first difference $\rm\:f(n\!+\!1)-f(n)\:$ is easy using the factored form of $\rm\,f,\,$ viz.
$$\rm\begin{eqnarray}\rm f(n\!+\!1)-f(n) &\,=\,&\rm n^2(n\!+\!1)(n\!+\!2)-(n\!-\!1)^2n(n\!+\!1)\\  &=&\rm n(n\!+\!1)\,(n(n\!+\!2)-(n\!-\!1)^2)\\ &=&\rm n(n\!+\!1)(4n\!-\!1)\end{eqnarray}$$
A: If you want to use induction, you are looking to assume $n^4-n^3-n^2+n$ is a multiple of $6$ and prove $(n+1)^4-(n+1)^3-(n+1)^2+n$ is.  So $$((n+1)^4-(n+1)^3-(n+1)^2+n+1)-(n^4-n^3-n^2+n)=4(n+1)^3+6(n+1)^2+4(n+1)+1-3(n+1)^2-3(n+1)-1-2(n+1)-1+1=4n^3+3n^2-n+=n(n+1)(4n-1)$$
One of the first two terms is even and one of the three is a multiple of $3$, so the difference $f(n+1)-f(n)$ is divisible by $6$
