2.102. $\left(\frac{(a-1)^{-1}}{a^{-3}}-(1-a)^{-1}\right) \cdot \frac{1+a(a-2)}{a^{2}-a+1} \cdot \sqrt{\frac{1}{(a+1)^{2}}}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

2.102. $\left(\frac{(a-1)^{-1}}{a^{-3}}-(1-a)^{-1}\right) \cdot \frac{1+a(a-2)}{a^{2}-a+1} \cdot \sqrt{\frac{1}{(a+1)^{2}}}$.

The above text is already in a mathematical expression and does not require translation. However, if you need the problem statement in English, it would be:

2.102. $\left(\frac{(a-1)^{-1}}{a^{-3}}-(1-a)^{-1}\right) \cdot \frac{1+a(a-2)}{a^{2}-a+1} \cdot \sqrt{\frac{1}{(a+1)^{2}}}$.

Simplify the expression above.
## Solution.

Domain of definition: $\left\{\begin{array}{l}a \neq 0, \\ a \neq \pm 1 .\end{array}\right.$

$$
\begin{aligned}
& \left(\frac{(a-1)^{-1}}{a^{-3}}-(1-a)^{-1}\right) \cdot \frac{1+a(a-2)}{a^{2}-a+1} \cdot \sqrt{\frac{1}{(a+1)^{2}}}=\left(\frac{\frac{1}{a-1}}{\frac{1}{a^{3}}}-\frac{1}{1-a}\right) \times \\
& \times \frac{1+a^{2}-2 a}{a^{2}-a+1} \cdot \frac{1}{|a+1|}=\left(\frac{a^{3}}{a-1}+\frac{1}{a-1}\right) \cdot \frac{a^{2}-2 a+1}{a^{2}-a+1} \cdot \frac{1}{|a+1|}=\frac{a^{3}+1}{a-1} \times \\
& \times \frac{(a-1)^{2}}{a^{2}-a+1} \cdot \frac{1}{|a+1|}=\frac{(a+1)\left(a^{2}-a+1\right)(a-1)}{\left(a^{2}-a+1\right) \cdot|a+1|}=\frac{(a+1)(a-1)}{|a+1|}= \\
& =\left\{\begin{array}{c}
\frac{(a+1)(a-1)}{-(a+1)}=1-a, \text { if } a+1<0, \text { or, considering the domain of definition, } \\
a<-1 ; \frac{(a+1)(a-1)}{a+1}=a-1, \text { if } a+1>0, \text { or, considering the domain of definition, } \\
a>-1, a \neq 0 \text { and } a \neq 1 .
\end{array}\right.
\end{aligned}
$$

Answer: $1-a$ for $a \in(-\infty ;-1) ; a-1$ for $a \in(-1 ; 0) \cup(0 ; 1)(1 ; \infty)$.