Question

Which is the simplified form of (9c^-9)^-3?

Answer 1

Answer:

We conclude that:

$\left(9c^{-9}\right)^{-3}=\frac{c^{27}}{729}$

Step-by-step explanation:

Given the expression

$\left(9c^{-9}\right)^{-3}$

Apply exponent rule:   $a^{-b}=\frac{1}{a^b}$

$\left(9c^{-9}\right)^{-3}=\frac{1}{\left(9c^{-9}\right)^3}$

Let us first solve:

$\left(9c^{-9}\right)^3$

Apply exponent rule:  $\left(a\cdot \:b\right)^n=a^nb^n$

$\left(9c^{-9}\right)^3=9^3\left(c^{-9}\right)^3$

             $=729\left(c^{-9}\right)^3$

Apply exponent rule:   $\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

             $=729c^{-9\cdot \:3}$

             $=729c^{-27}$

Apply exponent rule:   $a^{-b}=\frac{1}{a^b}$

             $=729\cdot \frac{1}{c^{27}}$

             $=\frac{729}{c^{27}}$

Therefore, the expression $\left(9c^{-9}\right)^{-3}=\frac{1}{\left(9c^{-9}\right)^3}$ becomes

$\left(9c^{-9}\right)^{-3}=\frac{1}{\left(9c^{-9}\right)^3}$

               $=\frac{1}{\frac{729}{c^{27}}}$         ∵  $\left(9c^{-9}\right)^3=\frac{729}{c^{27}}$

               $=\frac{c^{27}}{729}$          ∵  $\frac{1}{\frac{b}{c}}=\frac{c}{b}$

Hence, we conclude that:

$\left(9c^{-9}\right)^{-3}=\frac{c^{27}}{729}$