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?

$$  \left(   x^{6}  - 7 \right)  $$ 
$$  \left(   x^{2}  + 1 \right)  $$ 
$$  \left(   x^{6}  - 7 \right)  =  \left(   x^{2}  + 1 \right)  \cdot \color{magenta}{  \left(   x^{4}  -  x^{2}  + 1 \right) } +  \left( -8  \right)  $$ 
 $$  \left(   x^{2}  + 1 \right)  =  \left( -8  \right)  \cdot \color{magenta}{  \left(   \frac{  -  x^{2}  - 1 }{ 8 }  \right) } +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$ \color{magenta}{  \left(   x^{4}  -  x^{2}  + 1 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x^{4}  -  x^{2}  + 1 \right) }{ \left( 1  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{  -  x^{2}  - 1 }{ 8 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  -  x^{6}  + 7 }{ 8 }  \right) }{ \left(   \frac{  -  x^{2}  - 1 }{ 8 }  \right) } $$ 
 $$  \left(   x^{6}  - 7 \right)  \left( \frac{ 1}{8 } \right)  -  \left(   x^{2}  + 1 \right)  \left(   \frac{  x^{4}  -  x^{2}  + 1 }{ 8 }  \right)  =  \left( -1  \right)  $$ 

