Question

Problem in the comments below...

Answer 1

Fine the lowest common denominator. \[\frac{ 1 }{ (a+1)^2 } \times \frac{ 1 }{ (a-1)^2 } \times \frac{ 2 }{ a^2-1 }\]

Answer 2

you can use a difference of squares for the 3rd numerator.

Difference of squares. \(\large\color{slate}{ {\rm \color{red}{a}} ^2-{\rm \color{green}{b}}^2=( {\rm \color{red}{a}}-{\rm \color{green}{b}})( {\rm \color{red}{a}}+{\rm \color{green}{b}}) }\)

Answer 3

In your case, \(\large\color{slate}{ {\rm \color{red}{a}} ^2-{\rm \color{green}{1}}^2=? }\)

Answer 4

Weird in a multiplication problem, but okay. ^^ Difference of Squares is useful ^^

Answer 5

OK so where do I go from there?

Answer 6

what have you got for the 3rd denominator using a difference of squares ?

Answer 7

(a+1)(a-1)

Answer 8

yup

Answer 9

So, you have: \(\large\color{slate}{ \displaystyle \frac{1}{(a+1)^2}+\frac{1}{(a-1)^2}+\frac{1}{(a+1)(a-1)} }\)

Answer 10

I am assuming that what you wrote as multiplication in your first reply, was supposed to be +'s.

Answer 11

or, whether it is + or \(\large\color{slate}{ \times }\) doesn't mater much here.

Answer 12

No it's supposed to be multiplication.

Answer 13

oh, either way...

Answer 14

In general, \(\large\color{slate}{ \displaystyle \frac{1}{A^2} ~\times~\frac{1}{B^2}~\times~\frac{1}{AB} }\) will have a

common denominator of \(\large\color{slate}{ \displaystyle A^2B^2 }\)

Answer 15

So how do I apply that to the current equation?