Let's simplify the following expression:

$$
\frac{1-a^{2}}{\left(\frac{1-a \sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a \sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)}-1
$$
I. solution. The first factor of the denominator can be written as:

$$
\frac{1-a \sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}=\frac{(1-a)+\sqrt{a}(1-a)}{1-\sqrt{a}}=\frac{(1-a)(1+\sqrt{a})}{1-\sqrt{a}}
$$

Similarly, the second factor is:

$$
\frac{(1-a)(1-\sqrt{a})}{1+\sqrt{a}}
$$

thus the denominator is $(1-a)^{2}$. Therefore, the entire expression is

$$
\frac{1-a^{2}-(1-a)^{2}}{(1-a)^{2}}=\frac{2 a(1-a)}{(1-a)^{2}}=\frac{2 a}{1-a}
$$

Of course, we assumed that $a \neq 1$, otherwise the given expression is undefined, but the simplified final result is also undefined.

Szongoth Gábor (Budapest, I. István g. II. o. t.)

II. solution. Notice that $a \sqrt{a}=(\sqrt{a})^{3}$. Thus, based on the identity

$$
\frac{1-b^{3}}{1-b}=1+b+b^{2}
$$

- first using $b=\sqrt{a}$, then $b=-\sqrt{a}$ - the two factors of the denominator are

$$
1+2 \sqrt{a}+a=(1+\sqrt{a})^{2}, \quad \text { or } \quad 1-2 \sqrt{a}+a=(1-\sqrt{a})^{2}
$$

so the denominator, using the identity $c^{2} d^{2}=(c d)^{2}$, is $(1-a)^{2}$. We can proceed as in the I. solution.