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?)

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 $6$
