1=\frac{\left(x-1\right)^{2}\left(x+1\right)}{\ln\left(1-x\right)}
You asked:
Solve the equation \(1 = \frac{{\left( x - 1 \right)}^{2} \cdot \left(x + 1\right)}{\ln\left( 1 - x \right)}\) for the variable \(x\).
MathBot Answer:
The complex solutions are: \[\left\{x\; \middle|\; x \in \mathbb{R} \wedge x^{3} - x^{2} - x - \log{\left(1 - x \right)} + 1 = 0 \right\} \setminus \left\{0\right\}\]