The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-16&81&-126\\& & 3& -39& \color{black}{126} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{42}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-16x^{2}+81x-126 }{ x-3 } = \color{blue}{x^{2}-13x+42} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&81&-126\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-16&81&-126\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&81&-126\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 3 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -16 }&81&-126\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&81&-126\\& & 3& \color{blue}{-39} & \\ \hline &1&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 81 } + \color{orangered}{ \left( -39 \right) } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrr}3&1&-16&\color{orangered}{ 81 }&-126\\& & 3& \color{orangered}{-39} & \\ \hline &1&-13&\color{orangered}{42}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 42 } = \color{blue}{ 126 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&81&-126\\& & 3& -39& \color{blue}{126} \\ \hline &1&-13&\color{blue}{42}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -126 } + \color{orangered}{ 126 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&-16&81&\color{orangered}{ -126 }\\& & 3& -39& \color{orangered}{126} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{42}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-13x+42 } $ with a remainder of $ \color{red}{ 0 } $.