Using the Zero Exponent Rule

Use the zero exponent and other rules to simplify each expression.

ⓐ
$\begin{array}{ccc}\hfill \frac{c^3}{c^3}& =& c^{3-3}\hfill \\ & =& c^0\hfill \\ & =& 1\hfill \end{array}$

ⓑ
$\begin{array}{ccc}\hfill \frac{\mathrm{-3}{x}^5}{x^5}& =& \mathrm{-3}\cdot \frac{x^5}{x^5}\hfill \\ & =& \mathrm{-3}\cdot x^{5-5}\hfill \\ & =& \mathrm{-3}\cdot x^0\hfill \\ & =& \mathrm{-3}\cdot 1\hfill \\ & =& \mathrm{-3}\hfill \end{array}$

ⓒ
$\begin{array}{cccc}\hfill \frac{{\left(j^{2}k\right)}^4}{\left(j^{2}k\right)\cdot {\left(j^{2}k\right)}^3}& =& \frac{{\left(j^{2}k\right)}^4}{{\left(j^{2}k\right)}^{1+3}}\hfill & \text{Use the product rule in the denominator}.\hfill \\ & =& \frac{{\left(j^{2}k\right)}^4}{{\left(j^{2}k\right)}^4}\hfill & \text{Simplify}.\hfill \\ & =& {\left(j^{2}k\right)}^{4-4}\hfill & \text{Use the quotient rule}.\hfill \\ & =& {\left(j^{2}k\right)}^0\hfill & \text{Simplify}.\hfill \\ & =& 1\hfill & \end{array}$

ⓓ
$\begin{array}{cccc}\hfill \frac{5{\left(r{s}^2\right)}^2}{{\left(r{s}^2\right)}^2}& =& 5{\left(r{s}^2\right)}^{2-2}\hfill & \text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^2\right)}^0\hfill & \text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill & \text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill & \text{Simplify}.\hfill \end{array}$