The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&-6&11&-6\\& & 1& -5& \color{black}{6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The remainder when $ x^{3}-6x^{2}+11x-6 $ is divided by $ x-1 $ is $ \, \color{red}{ 0 } $.
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-6&11&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&-6&11&-6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-6&11&-6\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 1 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ -6 }&11&-6\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-6&11&-6\\& & 1& \color{blue}{-5} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}1&1&-6&\color{orangered}{ 11 }&-6\\& & 1& \color{orangered}{-5} & \\ \hline &1&-5&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&-6&11&-6\\& & 1& -5& \color{blue}{6} \\ \hline &1&-5&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&1&-6&11&\color{orangered}{ -6 }\\& & 1& -5& \color{orangered}{6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.