The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&0&-7&-6\\& & 3& 9& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-7x-6 }{ x-3 } = \color{blue}{x^{2}+3x+2} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&-7&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&0&-7&-6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&-7&-6\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 0 }&-7&-6\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&-7&-6\\& & 3& \color{blue}{9} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 9 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}3&1&0&\color{orangered}{ -7 }&-6\\& & 3& \color{orangered}{9} & \\ \hline &1&3&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&-7&-6\\& & 3& 9& \color{blue}{6} \\ \hline &1&3&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&0&-7&\color{orangered}{ -6 }\\& & 3& 9& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+3x+2 } $ with a remainder of $ \color{red}{ 0 } $.