Answer
$\frac{16a^2-12ab-18b^2}{(3a+4b)(3a-4b)(2a-b)}$.
Work Step by Step
The given expression is
$=\frac{6a+5b}{6a^2+5ab-4b^2}-\frac{a+2b}{9a^2-16b^2}$
First denominator
$=6a^2+5ab-4b^2$.
Rewrite the middle term $5ab$ as $8ab-3ab$
$=6a^2+8ab-3ab-4b^2$.
Group terms.
$=(6a^2+8ab)+(-3ab-4b^2)$.
Factor each term.
$=2a(3a+4b)-b(3a+4b)$.
Factor out $(3a+4b)$.
$=(3a+4b)(2a-b)$.
Second denominator $=9a^2-16b^2$.
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(3a)^2-(4b)^2$
$=(3a+4b)(3a-4b)$
Substitute all factors into the given expression.
$=\frac{6a+5b}{(3a+4b)(2a-b)}-\frac{a+2b}{(3a+4b)(3a-4b)}$
The LCM of all the denominators is
$=(3a+4b)(3a-4b)(2a-b)$.
Multiply all fractions to make denominators equal.
$=\frac{6a+5b}{(3a+4b)(2a-b)}\times \frac{(3a-4b)}{(3a-4b)}-\frac{a+2b}{(3a+4b)(3a-4b)}\times \frac{(2a-b)}{(2a-b)}$
Simplify.
$=\frac{(6a+5b)(3a-4b)}{(3a+4b)(3a-4b)(2a-b)}-\frac{(a+2b)(2a-b)}{(3a+4b)(3a-4b)(2a-b)}$
$=\frac{(6a+5b)(3a-4b)-(a+2b)(2a-b)}{(3a+4b)(3a-4b)(2a-b)}$
$=\frac{18a^2-24ab+15ab-20b^2-2a^2+ab-4ab+2b^2}{(3a+4b)(3a-4b)(2a-b)}$
Simplify.
$=\frac{16a^2-12ab-18b^2}{(3a+4b)(3a-4b)(2a-b)}$.
