Solution:

$\rightarrow$ Here, $f(x)=x^{3}-x-1=0$

at $x=1$, f(i) $=x-x-1=-1 \quad(-v e)$

$ x=2, \quad f(2)=8-2-1=5(+v e) \\ $

$\therefore$ The root lies between 1 and 2

$$ \begin{array}{|c|c|c|c|c|} \hline \text { Sr.No. } & \text { Negative } & \text { positive } & x_{i}=\frac{4+b}{2} & f(x i) \\ & \text { value } & \text { value } & & \\ \hline 1 & 1 & 2 & 1.5 & 0.875\gt0 \\\\ \hline 2 & 1 & 1.5 & 1.25 & -0.296875\lt0 \\\\ \hline 3 & 1.25 & 1.5 & 1.375 & 0.22461\gt0 \\\\ \hline 4 & 1.25 & 1.375 & 1.3125 & -0.0515\lt0 \\\\ \hline 5 & 1.3125 & 1.375 & 1.34375 & 0.0826\gt0 \\\\ \hline 6 & 1.3125 & 1.34375 & 1.328125 & 0.014576\gt0 \\\\ \hline 7 & 1.3125 & 1.328125 & 1.32031 & -0.0187\lt0 \\\\ \hline 8 & 1.32031 & 1.328125 & 1.32422 & -0.002122\lt0 \\\\ \hline 9 & 1.32422 & 1.328125 & 1.3262 & 0.006329\gt0 \\\\ \hline 10 & 1.32422 & 1.3262 & 1.3252 & 0.002056\gt0 \\\\ \hline 11 & 1.32422 & 1.3252 & 1.32471 & -0.000039\lt0 \\\\ \hline 12 & 1.32471 & 1.3252 & 1.325 & 0.00120\gt0 \\\\ \hline 13 & 1.325 & 1.3252 & 1.325 & \\\\ \hline \end{array} $$

Hence the correct root up to three decimal places is 1.325