Given Information:

The given expression is

$\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{{c}^{2}\text{}-\text{}9}\text{}+\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}-\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}$
Formula:

The algebraic identity used is

${a}^{2}\text{}-\text{}{b}^{2}=(a\text{}+\text{}b)(a\text{}-\text{}b).$
Calculation:

Consider the expression,

$\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{{c}^{2}\text{}-\text{}9}\text{}+\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}-\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}$
First, factor the denominators of this expression and apply

${a}^{2}\text{}-\text{}{b}^{2}=(a\text{}+\text{}b)(a\text{}-\text{}b)\text{}in\text{}{c}^{2}\text{}-\text{}9$ as follows:

$\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{{c}^{2}\text{}-\text{}9}\text{}+\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}-\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}=\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{(c\text{}-\text{}3)(c\text{}+\text{}3)}\text{}+\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}-\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}$
Now, consider the denominators,

$(c\text{}-\text{}3)(c\text{}+\text{}3),\text{}c\text{}+\text{}3\text{}and\text{}c\text{}-\text{}3.$
The least common denominator

$(LCD)\text{}is\text{}(c\text{}-\text{}3)(c\text{}+\text{}3).$
Now multiply the numerator and the denominator of

$\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{(c\text{}-\text{}3)(c\text{}+\text{}3)}\text{}with\text{}1,\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}with\text{}c\text{}-\text{}3\text{}and\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}\text{}with\text{}c\text{}+\text{}3$ to get the LCD in the denominator as follows:

$\frac{{c}^{2}\text{}+\text{}13c\text{}+\text{}18}{(c\text{}-\text{}3)(c\text{}+\text{}3)}\text{}+\text{}\frac{c\text{}+\text{}1}{c\text{}+\text{}3}\text{}-\text{}\frac{c\text{}+\text{}8}{c\text{}-\text{}3}=\text{}\frac{({c}^{2}\text{}+\text{}13c\text{}+\text{}18)(1)}{(c\text{}-\text{}3)(c\text{}+\text{}3)(1)}\text{}+\text{}\frac{(c\text{}+\text{}1)(c\text{}-\text{}3)}{(c\text{}+\text{}3)(c\text{}-\text{}3)}\text{}-\text{}\frac{(c\text{}+\text{}8)(c\text{}+\text{}3)}{(c\text{}-\text{}3)(c\text{}+\text{}3)}$
$=\frac{({c}^{2}\text{}+\text{}13c\text{}+\text{}18)(1)\text{}+\text{}(c\text{}+\text{}1)(c\text{}-\text{}3)\text{}-\text{}(c\text{}+\text{}8)(c\text{}+\text{}3)}{(c\text{}-\text{}3)(c\text{}+\text{}3)}$
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