Answer
$x^{2}+7x+5+\displaystyle \frac{8}{x-1}$
Work Step by Step
$\begin{array}{ccccccccccc}
& &x^{2} & +7x& +5 & \\
& &--&-- &--& \\
x-1&) & x^{3} & +6x^{2} & -2x& +3 & & \\
& & x^{3} & -x^{2 }& & & \color{red}{\leftarrow \small{x^{2}(x-1) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & 7x^{2} & -2x & +3 & \\
& & & 7x^{2} & -7x& & \color{red}{\leftarrow \small{7x(x-1) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & 5x & +3 & \\
& & & & 5x & -5 & \color{red}{\leftarrow \small{5(x-1) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & 8 &
\end{array}$
Quotient = $x^{2}+7x+5$
Remainder = $ 8$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ x^{3}+6x^{2}-2x+3}{x-1}$ = $x^{2}+7x+5+\displaystyle \frac{8}{x-1}$
