Chapter 5.6, Problem 32WE

To determine

To simplify the given expression.

Answer to Problem 32WE

z $\frac{4xy}{{\left(x-y\right)}^2{\left(x+y\right)}^2}$

Explanation of Solution

Given:

Expression: ${\left(x-y\right)}^{-2}-{\left(x+y\right)}^{-2}$

Calculation:

Consider the given expression ${\left(x-y\right)}^{-2}-{\left(x+y\right)}^{-2}$ ,

It can also be written as $\frac{1}{{\left(x-y\right)}^2}-\frac{1}{{\left(x+y\right)}^2}$ .

Since the denominators of both the fractions are different, find the least common denominator for both the fractions.

Here, least common denominator is ${\left(x-y\right)}^2{\left(x+y\right)}^2$ , so

  $\frac{1}{{\left(x-y\right)}^2}-\frac{1}{{\left(x+y\right)}^2}$

  $=\frac{1\cdot {\left(x+y\right)}^2}{{\left(x-y\right)}^2\cdot {\left(x+y\right)}^2}-\frac{1\cdot {\left(x-y\right)}^2}{{\left(x+y\right)}^2\cdot {\left(x-y\right)}^2}$

  $=\frac{{\left(x+y\right)}^2}{{\left(x-y\right)}^2{\left(x+y\right)}^2}-\frac{{\left(x-y\right)}^2}{{\left(x-y\right)}^2{\left(x+y\right)}^2}$

Now, as the denominators of the two fractions are same, simply subtract the numerators and put the result on the same denominator.

  $=\frac{{\left(x+y\right)}^2-{\left(x-y\right)}^2}{{\left(x-y\right)}^2{\left(x+y\right)}^2}$

  $\begin{array}{l}=\frac{\left(x^2+y^2+2xy\right)-\left(x^2+y^2-2xy\right)}{{\left(x-y\right)}^2{\left(x+y\right)}^2}\\ =\frac{x^2+y^2+2xy-x^2-y^2+2xy}{{\left(x-y\right)}^2{\left(x+y\right)}^2}\\ =\frac{4xy}{{\left(x-y\right)}^2{\left(x+y\right)}^2}\end{array}$

Conclusion:

Therefore, after simplification, the answer is $\frac{4xy}{{\left(x-y\right)}^2{\left(x+y\right)}^2}$ .