Answer
$ \displaystyle \frac{4b}{4b-3},\quad x\neq\frac{9}{4}$
Work Step by Step
$\displaystyle \frac{6b^{2}-10b}{16b^{2}-48b+27}+\frac{7b^{2}-20b}{16b^{2}-48b+27}-\frac{6b-3b^{2}}{16b^{2}-48b+27}$
... To add/subtract rational expressions with the same denominator,
add/subtract numerators and place the sum/difference over the common denominator.
$= \displaystyle \frac{6b^{2}-10b+(7b^{2}-20b)- (6b-3b^{2})}{16b^{2}-48b+27}$
$= \displaystyle \frac{6b^{2}-10b+7b^{2}-20b-6b+3b^{2}}{16b^{2}-48b+27}$
$= \displaystyle \frac{16b^{2}-36b}{16b^{2}-48b+27}$
Factor what we can,
... $16b^{2}-36b=4b(4b-9)$
... for $ax^{2}+bx+c$ we search for factors of $ac$ whose sum is $b$
... if sucessful, rewrite $bx $ and factor in pairs.
... here, factors of $16\cdot 27=2\times 2\times 2\times 2\times 3\times 3\times 3$
... whose sum is $-48$ are $-36=-2\times 2\times 3\times 3$ and $-12=-2\times 2\times 3$
$16b^{2}-48b+27=16b^{2}-36b-12b+27$
$=4b(4b-9)-3(4b-9)$
$=(4b-9)(4b-3)$
$= \displaystyle \frac{4b(4b-9)}{(4b-9)(4b-3)}$
... common factors cancel,
=$ \displaystyle \frac{4b}{4b-3},\quad x\neq\frac{9}{4}$
