Answer
$\frac{ 3ab(b+a)}{ (2b+3a)(2b-3a)}$.
Work Step by Step
The given expression is
$=\frac{3a^{-1}+3b^{-1}}{4a^{-2}-9b^{-2}}$
Multiply the numerator and the denominator by $a^2b^2$.
$=\frac{a^2b^2}{a^2b^2}\cdot \frac{3a^{-1+}3b^{-1}}{4a^{-2}-9b^{-2}}$
Use the distributive property.
$=\frac{a^2b^2 \cdot 3a^{-1}+a^2b^2 \cdot 3b^{-1}}{a^2b^2 \cdot 4a^{-2}-a^2b^2 \cdot 9b^{-2}}$
Simplify.
$=\frac{ 3a^{-1+2}b^2+ 3a^2b^{-1+2}}{ 4a^{-2+2}b^2- 9a^2b^{-2+2}}$
$=\frac{ 3a^{1}b^2+ 3a^2b^{1}}{ 4a^{0}b^2- 9a^2b^{0}}$
$=\frac{ 3ab^2+ 3a^2b}{ 4b^2- 9a^2}$
Factor the fraction.
Numerator $=3ab^2+3a^2b$.
Factor out the common terms.
$=3ab(b+a)$
Denominator $=4b^2-9a^2$.
$=(2b)^2-(3a)^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(2b+3a)(2b-3a)$.
Substitute all factors into the fraction.
$=\frac{ 3ab(b+a)}{ (2b+3a)(2b-3a)}$.
