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I'm in doubt on how to simplify $ \left (-\dfrac{1}{243} \right )^{-\frac{2}{3}}$.

I've started with $\left (-\dfrac{1}{9\sqrt{3}} \right )^{-\frac{2}{3}}$ but now I'm stuck because of this minus signal in the main fraction ?

Thanks in advance.

3 Answers 3

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$\left(-{1\over 243}\right)^{-2/3} = {\left(3^5\right)^{-2/3}} = 3^{10/3} = 27\root{3}\of 3$

You ditch the - because of the even power. This is dangerous and dicey tho' because of certain bad behavior between fractional powers and negative numbers.

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    That doesn't mean /you/ should. The point of this site is to clarify what regular texts don't.2012-06-25
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$ \left( -\frac{1}{243} \right)^{-\frac{2}{3}} = (-243)^\frac{2}{3} = \sqrt[3]{(-243)^2} = 27\sqrt[3]{3} $

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Take it one step at a time, first getting rid of the negative exponent by taking reciprocals:

$\begin{align*} \left(-\frac1{243}\right)^{-\frac23}&=\left(-\frac1{243}\right)^{(-1)\cdot\frac23}\\ &=\left(\left(-\frac1{243}\right)^{-1}\right)^{\frac23}\\ &=(-243)^{2/3}\\ &=\big(-(3^5)\big)^{2/3}\\ &=\big(-(3^5)\big)^{2\cdot\frac13}\\ &=\left(\big(-(3^5)\big)^2\right)^{1/3}\\ &=\left(3^{10}\right)^{1/3}\\ &=3^{10/3}\\ &=3^{3+\frac13}\\ &=3^3\cdot3^{1/3}\\ &=27\sqrt[3]3\;. \end{align*}$

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    @robjohn: Beats me; I apparently concentrated so hard on the reciprocal that I forgot the minus sign altogether.2012-06-24