Find the quotient and the remainder of $(n^6-7)/(n^2+1)$ 
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.

So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?

Yet another way:
$$
\begin{align}
n^6-7 &= \big(\left(n^2\color{red}{+1}\right)\color{red}{-1}\big)^3-7 \\
 &= \left(n^2+1\right)^3-3\left(n^2+1\right)^2+3\left(n^2+1\right)-1 -7 \\
 &= \left(n^2+1\right)\big(\left(n^2+1\right)^2-3\left(n^2+1\right)+3\big) - 8 \\
 &= \left(n^2+1\right)\left(n^4-n^2+1\right) - 8
\end{align}
$$
