Question
Simplify and express each as a rational number: $\Big(\frac{-7}{8}\Big)^{-3}\times \Big(\frac{-7}{8}\Big)^2$
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Answer
$\Big(\frac{-7}{8}\Big)^{-3}\times \Big(\frac{-7}{8}\Big)^2=\Big(\frac{-7}{8}\Big)^{-3+2}$ $=\Big(\frac{-7}{8}\Big)^{-1}=\Big(\frac{8}{-7}\Big)^1=\frac{8}{-7}$ $=\frac{8\times(-1)}{-7\times(-1)}$ $=\frac{-8}{7}$
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