## Related questions with answers

Simplify. Suppose that all variables represent nonzero integers.

$\left[\left(\frac{a^{-2 c}}{b^{7 c}}\right)^{-3}\left(\dfrac{a^{4 c}}{b^{-3 c}}\right)^2\right]^{-a}$

Solution

VerifiedUsing the laws of exponents and order of operations, the expression $\left[ \left( \dfrac{a^{-2c}}{b^{7c}} \right)^{-3}\left( \dfrac{a^{4c}}{b^{-3c}} \right)^{2} \right]^{-a}$ simplifies to

$\begin{align*} & \left[ \left( \dfrac{a^{-2c(-3)}}{b^{7c(-3)}} \right)\left( \dfrac{a^{4c(2)}}{b^{-3c(2)}} \right) \right]^{-a} \\\\&= \left[ \left( \dfrac{a^{6c}}{b^{-21c}} \right)\left( \dfrac{a^{8c}}{b^{-6c}} \right) \right]^{-a} \\\\&= \left[ \dfrac{a^{6c+8c}}{b^{-21c+(-6c)}} \right]^{-a} \\\\&= \left[ \dfrac{a^{14c}}{b^{-27c}} \right]^{-a} \\\\&= \dfrac{a^{14c(-a)}}{b^{-27c(-a)}} \\\\&= \dfrac{a^{-14ac}}{b^{27ac}} \\\\&= \dfrac{1}{a^{14ac}b^{27ac}} .\end{align*}$

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