Finding my x!

We have , $\frac{(x-3)^{2}-4}{x-1} =0 , \text{where } x \neq 1$ which means, $(x-3)^{2} - 4 =0 \Rightarrow (x-3)^2 = 4$ $x-3 = \pm 2 \Rightarrow x= 3\pm 2$ which gives, $x=5,1$ But $x \neq 1$ because then $x-1$ will be equal to $0$ making the expression in and indeterminate form. So , $x=\boxed {5}$

Here is my elementary solution:

$\frac{(x-3)^2-4}{x-1}=0$

$\frac{(x-3)^2-2^2}{x-1}=0$

$\frac{(x-3-2)(x-3+2)}{x-1}=0$

$\Rightarrow x-5=0 \Rightarrow x=\boxed{5}$

((x-3)^2-2^2)/(x-1)=0 =or,((x-5)(x-1))/(x-1)=o or,x=5