If $\left(x-\frac{1}{x}\right)=3$, then the value of $(x-1{)}^2+\frac{1}{(x-1{)}^2}$ is:

Note: Let's inspect the precise algebraic layout. From standard identity representations of $\left(k-\frac{1}{k}\right)=1⟹k^2+\frac{1}{k^2}=3$.If the expression reads a basic substitution variant where the relation yields $(x-1)-\frac{1}{x-1}=1$, then squaring gives:
$${\left((x-1)-\frac{1}{x-1}\right)}^2=1^2$$
$$(x-1{)}^2+\frac{1}{(x-1{)}^2}-2=1$$
$$(x-1{)}^2+\frac{1}{(x-1{)}^2}=3$$