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Simplify: \[(3z)^{2} (6z^{2})^{-3} \]
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\[(3z ^{2})(6z ^{2})^{-3}=(3z ^{2})6^{-3} z ^{-6}=3*6^{-3}*z ^{-4}=z ^{-4}/108\]
answer is \[{1 \over 24z^{4}}\]
sorry your prob was (3z)^2...and i did it with 3z^2....so i got wrong...u r write..
can you explain how to get the answer?
\[(3z)^2(6z^2)^{-3}=(9z^2)/(6z^2)^3=\frac{9z^2}{216z^6}=1/24z^4\]
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okay. okay... i'm still trying to grasp this
wait... it's not making sense. why divide on the second step?
where did you get 216z^6 on the third step?
a^-x=1/(a^x) do u know that rule?
nope. haa
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(6z^2)^3=6^3z^6=216z^6
okay i see where the divde comes from. the negative exponent.
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