Question #1bb1d

Dec 20, 2017

See a solution process below:

Explanation:

First, excluded values would be when the denominator is equal to $0$ or:

$10 {h}^{2} {z}^{3} = 0$ or when $h = 0$ or $z = 0$

First, rewrite the expression as:

$\frac{3 \times 2}{5 \times 2} \left({h}^{3} / {h}^{2}\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{5 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \left({h}^{3} / {h}^{2}\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3}{5} \left({h}^{3} / {h}^{2}\right) \left(\frac{z}{z} ^ 3\right)$

Next, use these rules of exponents to simplify the $h$ terms:

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

$\frac{3}{5} \left({h}^{\textcolor{red}{3}} / {h}^{\textcolor{b l u e}{2}}\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3}{5} \left({h}^{\textcolor{red}{3} - \textcolor{b l u e}{2}}\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3}{5} \left({h}^{1}\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3}{5} \left(h\right) \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3 h}{5} \left(\frac{z}{z} ^ 3\right)$

Now, use these rules of exponents to simplify the $z$ term:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$\frac{3 h}{5} \left(\frac{z}{z} ^ 3\right) \implies$

$\frac{3 h}{5} \left({z}^{\textcolor{red}{1}} / {z}^{\textcolor{b l u e}{3}}\right) \implies$

$\frac{3 h}{5} \left(\frac{1}{z} ^ \left(\textcolor{b l u e}{3} - \textcolor{red}{1}\right)\right) \implies$

$\frac{3 h}{5} \left(\frac{1}{z} ^ 2\right) \implies$

$\frac{3 h}{5 {z}^{2}}$