<s>[INST] Solve $x=\sqrt{x-x^{-1}}+\sqrt{1-x^{-1}}$ Find all real numbers of $x$ such that $$x=\left(x-\frac 1x\right)^{\frac 12}+\left(1-\frac 1x\right)^{\frac 12}$$
My Attempt: Square both sides to get$$x^2=x-\frac 1x+1-\frac 1x+2\sqrt{\left(x-\frac 1x\right)\left(1-\frac 1x\right)}$$
Then move all the terms to the left side except for the square root and then square the equation. But I predict the terms to get really ugly really fast.
So I'm wondering if there is an elegant way of obtaining the solution without going through much of a hassle.
 [/INST] Multiply the original equation 
$$
(x-x^{-1})^{1/2}+(1-x^{-1})^{1/2}=x\tag{1}
$$
by $(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}$ to get
$$
x-1=x((x-x^{-1})^{1/2}-(1-x^{-1})^{1/2})
$$
That is
$$
(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}=1-x^{-1}\tag{2}
$$
Denote $a=(x-x^{-1})^{1/2}$ and $b=(1-x^{-1})^{1/2}$. Then $(1)$ and $(2)$ can be rewritten as 
$$
a+b=x\tag{3}
$$
$$ 
a-b=b^2\tag{4}
$$ 
From the very definition of $a$ and $b$ we have $a^2-b^2=x-1$. With $(3)$ we get $$
a^2-b^2=a+b-1\tag{5}
$$ 
Subtracting $(4)$ from $(5)$ we obtain $a^2-a=a-1$. Hence $a=1$. The rest is clear.
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