Chapter A.3, Problem 63E

Solution

Simplify the compound fraction.

  $\frac{(x^2+x\cdot y+y^2)}{(x+y)}$ .

Given:

The given compound fraction is $\frac{\frac{x}{y^2}-\frac{y}{x^2}}{\frac{1}{y^2}-\frac{1}{x^2}}$ .

Concept Used:

• Compound fraction is a fraction with fractions in the numerator or denominator.
• First multiply the denominator of the fraction by the whole number. 
• Then, add that number to the numerator of the fraction, and keep the original denominator the same.

Calculation:

The given compound fraction is $\frac{\frac{x}{y^2}-\frac{y}{x^2}}{\frac{1}{y^2}-\frac{1}{x^2}}$ .

  $\begin{array}{c}\frac{\frac{x}{y^2}-\frac{y}{x^2}}{\frac{1}{y^2}-\frac{1}{x^2}}=\frac{\frac{x^3-y^3}{y^2{x}^2}}{\frac{x^2-y^2}{y^2{x}^2}}\\ =\frac{y^2{x}^2(x^3-y^3)}{y^2{x}^2(x^2-y^2)}\\ =\frac{(x^3-y^3)}{(x^2-y^2)}\\ =\frac{(x-y)(x^2+x\cdot y+y^2)}{(x+y)(x-y)}\\ =\frac{(x^2+x\cdot y+y^2)}{(x+y)}\end{array}$

Thus by simplifying the compound fraction $\frac{\frac{x}{y^2}-\frac{y}{x^2}}{\frac{1}{y^2}-\frac{1}{x^2}}$ the solution is $\frac{(x^2+x\cdot y+y^2)}{(x+y)}$ .