The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&-3&0&-108\\& & 3& 0& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-108} \end{array} $$Because the remainder $ \left( \color{red}{ -108 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 3x^{3}-3x^{2}-108$.
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-3&0&-108\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&-3&0&-108\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-3&0&-108\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ -3 }&0&-108\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-3&0&-108\\& & 3& \color{blue}{0} & \\ \hline &3&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&3&-3&\color{orangered}{ 0 }&-108\\& & 3& \color{orangered}{0} & \\ \hline &3&0&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-3&0&-108\\& & 3& 0& \color{blue}{0} \\ \hline &3&0&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -108 } + \color{orangered}{ 0 } = \color{orangered}{ -108 } $
$$ \begin{array}{c|rrrr}1&3&-3&0&\color{orangered}{ -108 }\\& & 3& 0& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{-108} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -108 }\right)$.