Answer
$\left\{\dfrac{2}{3},2\right\}$
Work Step by Step
Let $z=
(x-1)^{-1}
$. Then the given equation, $
3-2(x-1)^{-1}=(x-1)^{-2}
,$ is equivalent to
\begin{align*}\require{cancel}
3-2z&=z^2
\\
0&=z^2+2z-3
\\
z^2+2z-3&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+3)(z-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+3=0 & z-1=0
\\
z=-3 & z=1
.\end{array}
Since $z=
(x-1)^{-1}=\dfrac{1}{x-1}
,$ by back substitution, then
\begin{array}{l|r}
\dfrac{1}{x-1}=-3 & \dfrac{1}{x-1}=1
.\end{array}
By cross-multiplication and the properties of equality, the equations above are equivalent to
\begin{array}{l|r}
1(1)=-3(x-1) & 1(1)=1(x-1)
\\
1=-3x+3 & 1=x-1
\\
3x=3-1 & 1+1=x
\\
3x=2 & 2=x
\\\\
x=\dfrac{2}{3} & x=2
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\dfrac{2}{3}: & \text{If }x=2
\\\\
3-2\left(\dfrac{2}{3}-1\right)^{-1}\overset{?}=\left(\dfrac{2}{3}-1\right)^{-2} &
3-2(2-1)^{-1}\overset{?}=(2-1)^{-2}
\\\\
3-2\left(-\dfrac{1}{3}\right)^{-1}\overset{?}=\left(-\dfrac{1}{3}\right)^{-2} &
3-2(1)^{-1}\overset{?}=(1)^{-2}
\\\\
3-2\left(-\dfrac{3}{1}\right)^{1}\overset{?}=\left(-\dfrac{3}{1}\right)^{2} &
3-2(1)\overset{?}=1
\\\\
3+6\overset{?}=9 &
3-2\overset{?}=1
\\
9\overset{\checkmark}=9 &
1\overset{\checkmark}=1
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
3-2(x-1)^{-1}=(x-1)^{-2}
$ is $\left\{\dfrac{2}{3},2\right\}$.
