Step 1: The Regula Falsi formula comes from finding where the straight line (chord) joining \((x_0,f(x_0))\) and \((x_1,f(x_1))\) crosses the x-axis. The chord has equation
\[y-f(x_0) = \frac{f(x_1)-f(x_0)}{x_1-x_0}(x-x_0)\]Setting \(y=0\) and solving for \(x\) gives \(x = x_0 - f(x_0)\cdot\frac{x_1-x_0}{f(x_1)-f(x_0)}\), which is algebraically identical to the standard formula.
Step 2: With \(x_0=1\), \(f(x_0)=-1\), \(x_1=2\), \(f(x_1)=5\):
\[x_2 = 1-(-1)\cdot\frac{2-1}{5-(-1)} = 1+\frac{1}{6} = \frac{7}{6}\approx1.1667\]This matches the value found by the standard formula, confirming the chord based construction.
Step 3: Compute \(f(x_2)\) exactly using fractions: \(x_2=\frac{7}{6}\), so
\[f\left(\frac{7}{6}\right)=\frac{343}{216}-\frac{7}{6}-1=\frac{343-252-216}{216}=\frac{-125}{216}\approx-0.5787\]Step 4: Since this is still negative, the new chord joins \((x_2,f(x_2))=\left(\frac{7}{6},-\frac{125}{216}\right)\) and \((x_1,f(x_1))=(2,5)\). Its x-intercept is
\[x_3 = x_2-f(x_2)\cdot\frac{x_1-x_2}{f(x_1)-f(x_2)}\]Step 5: Substitute \(x_1-x_2 = 2-\frac{7}{6}=\frac{5}{6}\) and \(f(x_1)-f(x_2)=5+\frac{125}{216}=\frac{1205}{216}\):
\[x_3 = \frac{7}{6} - \left(-\frac{125}{216}\right)\cdot\frac{5/6}{1205/216} = \frac{7}{6}+\frac{125}{216}\cdot\frac{5}{6}\cdot\frac{216}{1205}\]Step 6: Simplify the fraction:
\[x_3 = \frac{7}{6} + \frac{125\times5}{6\times1205} = \frac{7}{6}+\frac{625}{7230} = \frac{7}{6}+\frac{125}{1446} = \frac{1687}{1446}+\frac{125}{1446} = \frac{1812}{1446} = \frac{302}{241}\]Converting to a decimal gives \(x_3 \approx 1.2531\).
Step 7: Rounded to two decimal places:
\[\boxed{x_3\approx1.25}\]