Question

(a^-1 + b^-1) / (a^-2 - b^-2) simplify​

Answer 1

We must recall that:

In mathematics, an identity element, or neutral element, of a binary operation operating on a set is an element of the set which leaves unchanged every element of the set when the operation is applied.

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

$\frac{(a+b)(a^2b^2)}{(ab)(a^2-b^2)}=\frac{(a+b)(ab)}{a^2-b^2}$

Use identify, $a^2-b^2=(a+b)(a-b)$

$\frac{-(a+b)(ab)}{(a+b)(a-b)}=\frac{ab}{a-b}$

Answer 2

Given: Here we have given $\frac{{{a}^{-1}}+{{b}^{-1}}}{{{a}^{-2}}-{{b}^{-2}}}$

To find: we have find the value of $\frac{{{a}^{-1}}+{{b}^{-1}}}{{{a}^{-2}}-{{b}^{-2}}}$

Solution:

Here we have the given

$\frac{{{a}^{-1}}+{{b}^{-1}}}{{{a}^{-2}}-{{b}^{-2}}}=\frac{\frac{1}{a}+\frac{1}{b}}{{{\left( \frac{1}{a} \right)}^{2}}-{{\left( \frac{1}{b} \right)}^{2}}}$

We will use the identity

${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

$\begin{align} & =\frac{\frac{1}{a}+\frac{1}{b}}{\left( \frac{1}{a}+\frac{1}{b} \right)\left( \frac{1}{a}-\frac{1}{b} \right)} \\ & =\frac{1}{\left( \frac{1}{a}-\frac{1}{b} \right)} \\ & =\frac{1}{{{a}^{-1}}-{{b}^{-1}}} \\ \end{align}$

Final answer:

Hence the answer is $& \dfrac{1}{{{a}^{-1}}-{{b}^{-1}}} \\ \end{align}$