Chapter 6.2, Problem 53WE

To determine

To calculate: The simplified form of the expression ${\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}$.

Answer to Problem 53WE

The simplified form of the expression ${\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}$ is $\frac{\left(x-y\right)\left(x^2+y^2\right)}{{\left(x+y\right)}^3}$.

Explanation of Solution

Given information:

The expression ${\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}$.

Formula used:

Law of exponent for a quotient raised to power m where m is a positive integer is ${\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}$.

Fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Calculation:

Consider the expression ${\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}$.

Recall that law of exponent for a quotient raised to power m where m is a positive integer is ${\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}$.

Apply it,

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

Recall that difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Apply it,

  $\begin{array}{c}{\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}=\frac{{\left(x-y\right)}^2}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{x^2-y^2}\\ =\frac{{\left(x-y\right)}^2}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\\ =\frac{\left(x-y\right)}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{\left(x+y\right)}\end{array}$

Recall that fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Apply it,

  $\begin{array}{c}{\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}=\frac{{\left(x-y\right)}^2}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{x^2-y^2}\\ =\frac{{\left(x-y\right)}^2}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\\ =\frac{\left(x-y\right)}{{\left(x+y\right)}^2}\cdot \frac{x^2+y^2}{\left(x+y\right)}\\ =\frac{\left(x-y\right)\left(x^2+y^2\right)}{{\left(x+y\right)}^3}\end{array}$

Thus, the simplified form of the expression ${\left(\frac{x-y}{x+y}\right)}^2\cdot \frac{x^2+y^2}{x^2-y^2}$ is $\frac{\left(x-y\right)\left(x^2+y^2\right)}{{\left(x+y\right)}^3}$.