Write the following in exponential form assuming denominators not equal to zero:
$\frac{{(- 6)}^{8}}{{(- 6)}^{3}}$

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Solution

For any non-zero integer $a$ , where $m$ and $n$ are whole numbers and $m > n$ .
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$
So, $\frac{{(- 6)}^{8}}{{(- 6)}^{3}}$ $= {(- 6)}^{8 - 3}$ where $a = - 6, m = 8, n = 3$
$= {(- 6)}^{5}$ .
Therefore, $\frac{{(- 6)}^{8}}{{(- 6)}^{3}}$ $= {(- 6)}^{5}$ .

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