# How do you simplify (3a^2b^4+9a^3b-6a^5b)/(3a^2b) and what are the ecluded values of the variables?

Nov 5, 2017

See a solution process below:

#### Explanation:

First, the Excluded Values can be found by equating the denominator to $0$ and then solve each variable for $0$:

$2 {a}^{2} b = 0$

${a}^{2} = 0$ therefore $a = 0$

$b = 0$

The Excluded values are $a = 0$ and/or $b = 0$

To simplify the expression, rewrite the expression as:

$\frac{3 {a}^{2} {b}^{4}}{3 {a}^{2} b} + \frac{9 {a}^{3} b}{3 {a}^{2} b} - \frac{6 {a}^{5} b}{3 {a}^{2} b} \implies$

$\frac{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{3}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{{a}^{2}}}} {b}^{4}}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{3}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{{a}^{2}}}} b} + \frac{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{9}}} 3 {a}^{3} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{b}}}}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{3}}} {a}^{2} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{b}}}} - \frac{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{6}}} 2 {a}^{5} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{b}}}}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{3}}} {a}^{2} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{b}}}} \implies$

${b}^{4} / b + \frac{3 {a}^{3}}{a} ^ 2 - \frac{2 {a}^{5}}{a} ^ 2$

Next, use these rules for exponents to simplify the remaining terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${a}^{\textcolor{red}{1}} = a$

${b}^{\textcolor{red}{4}} / {b}^{\textcolor{b l u e}{1}} + \frac{3 {a}^{\textcolor{red}{3}}}{a} ^ \textcolor{b l u e}{2} - \frac{2 {a}^{\textcolor{red}{5}}}{a} ^ \textcolor{b l u e}{2} \implies$

${b}^{\textcolor{red}{4} - \textcolor{b l u e}{1}} + 3 {a}^{\textcolor{red}{3} - \textcolor{b l u e}{2}} - 2 {a}^{\textcolor{red}{5} - \textcolor{b l u e}{2}} \implies$

${b}^{3} + 3 {a}^{1} - 2 {a}^{3} \implies$

${b}^{3} + 3 a - 2 {a}^{3} \implies$