Question

In the following exercise, simplify$\frac{{\displaystyle \frac{3}{a^2}}-{\displaystyle \frac{1}{b}}}{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b^2}}}$

Short answer

The solution is$\frac{\left(3b-a^2\right)b}{a\left(b^2+a\right)}$

Step-by-step solution

Step 1. Given Information

$\frac{\left(3b-a^2\right)b}{a\left(b^2+a\right)}$

Step 2. Simplify

The LCD of all fractions in the complex rational expression is $\left(a^{2}b\right)\left(a{b}^2\right)$

Multiply the numerator and denominator by the LCD.

$$\begin{gathered} =\frac{\left(a^{2}b\right)\left(a{b}^2\right)\left({\displaystyle \frac{3}{a^2}}-{\displaystyle \frac{1}{b}}\right)}{\left(a^{2}b\right)\left(a{b}^2\right)\left({\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b^2}}\right)} \\ =\frac{\left({\displaystyle \left(a^{2}b\right)\left(a{b}^2\right)\frac{3}{a^2}-\left(a^{2}b\right)\left(a{b}^2\right)\frac{1}{b}}\right)}{\left({\displaystyle \left(a^{2}b\right)\left(a{b}^2\right)\frac{1}{a}+\left(a^{2}b\right)\left(a{b}^2\right)\frac{1}{b^2}}\right)} \\ =\frac{3a{b}^3-a^3{b}^2}{a^2{b}^3+a^{3}b} \\ =\frac{\left(3b-a^2\right)b}{a\left(b^2+a\right)} \end{gathered}$$