(64n^9)^^{2}/_{3 }(5b^2c^^{1}/_{4})^3
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Originally Posted by jobbles85 (64n^9)^^{2}/_{3 }(5b^2c^^{1}/_{4})^3 Not showing steps, that's your job. You need to use the following index laws: $\displaystyle \displaystyle \begin{align*} \left(a^m\right)^p &= a^{m \cdot p} \\ \\ \left(a \cdot b\right)^n &= a^n \cdot b^n \end{align*}$
Originally Posted by Prove It Not showing steps, that's your job. You need to use the following index laws: $\displaystyle \displaystyle \begin{align*} \left(a^m\right)^p &= a^{m \cdot p} \\ \\ \left(a \cdot b\right)^n &= a^n \cdot b^n \end{align*}$ so would the steps be.. =64n9x2/3 =64n^6
Originally Posted by jobbles85 so would the steps be.. =64n9x2/3 =64n^6 correction ... $\displaystyle (64n^9)^{\frac{2}{3}} = 64^{\frac{2}{3}}n^6$ finish by simplifying the factor $\displaystyle 64^{\frac{2}{3}}$ ...
(64n^9)^2/3 =642/3n^6 =(3√64^2)(n^6) =(3√4096)(n^6) =±16n^6 God I have no clue what I'm doing (5b^2c^1/4)^3 =(5)^3 (b^2)^3 (c^1/4)^3 =125b^5c^3/4
would have been really easy if you knew that $\displaystyle 64 = 4^3$ ... and it's not $\displaystyle \pm$ last one is correct
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