<s>[INST] Solve $ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$. Is my solution correct? Find the roots of the following equation, if any:
$$
\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
=
27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.
$$
My approach:
The following constraints should hold jointly for x:


*

*$1-x^2\geq0\iff x\in[-1,1]$

*$1+x\geq0 \iff x\geq-1$

*$1-x\geq0 \iff x\leq1$

*$\sqrt{1+x}\ne\sqrt{1-x}\iff x\ne0$


Consequently, $x\in[-1,0)\cup(0,1]$ and for such x's I solve and I reach at the following equation
$$
\left(\frac{x}{1+\sqrt{1-x^2}}\right)^3=27\implies
\frac{x}{1+\sqrt{1-x^2}}=3,
$$
which gives
$$
x = 3 + 3\sqrt{1-x^2}\implies
3\sqrt{1-x^2} = x-3\implies
9-9x^2=x^2-6x+9\implies
10x^2-6x=0\implies
x(10x-6)=0,
$$
and thus $x=0$ or $x=\frac{3}{5}$. The first one is rejected, but $x=\frac{3}{5}\in[-1,0)\cup(0,1]$.
However, if we plug $x=\frac{3}{5}$ into the original equation, we find out that this is impossible. So we should reject $x=\frac{3}{5}$. The original equation is impossible.
Is my approach correct? Thank you very much in advance.
 [/INST] The left-hand side is less than 1; the right-hand side is more than 27.
</s>