<s>[INST] Solving $ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$ 
Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$

Well I first observe that each factor has its own inverse involve in the equation and rewriting 
gives $$\left( \frac{x^2}{x-1}-\frac{x-1}{x^2} \right) + \left(\sqrt{x-1}  -\frac{1}{\sqrt{x-1}} \right)+\left(\frac{\sqrt{x-1}}{x^2} -\frac{x^2}{\sqrt{x-1}}\right) =0  $$
However after expanding all the terms it does not look wise to do so.  Does anyone has an idea  
 [/INST] Let $\sqrt {x-1}=t \implies x= 1+t^2 \quad x^2=(1+t^2)^2$
$$\frac{(1+t^2)^2}{t^2} + t +\frac{t}{(1+t^2)^2} = \frac{t^2}{(1+t^2)^2} + \frac{1}{t} + \frac{(1+t^2)^2}{t}$$
$$(1+t^2)^4 + t^3(1+t^2)^2 +t^3 = t^4 + t(1+t^2)^2 + t(1+t^2)^4$$
$$(t - 1)(t^2 - t + 1) (t^2 + t + 1) (t^4 + 2 t^2 - t + 1)=0 \implies t=1$$
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