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simplfy (3z)^2(6z^2)^-3
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\[3z^2 \times (6z^2)^{-3}\]\[3z^2/[(6z^2)^3]\]\[3z^2/(216z^6)\]\[1/(72z^4)\]
the answer in the book says \[1/24z ^{4}\]
\[\huge (3z)^2 \times (6z^2)^{-3}= \frac{(3z)^2}{(6z^2)^3}=\frac{9z^2}{216z^6}= \frac{1}{24z^4}\]
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