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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.

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up vote 2 down vote accepted

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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What happened to the minus sign in the first step? To me, the interesting part of this question is how a fractional power of a negative number should be handled. – robjohn Jun 24 '12 at 21:54
We harbor the same general feeling, robjohn. – ncmathsadist Jun 24 '12 at 21:59
@robjohn: Beats me; I apparently concentrated so hard on the reciprocal that I forgot the minus sign altogether. – Brian M. Scott Jun 24 '12 at 21:59

$$ \left( -\frac{1}{243} \right)^{-\frac{2}{3}} = (-243)^\frac{2}{3} = \sqrt[3]{(-243)^2} = 27\sqrt[3]{3} $$

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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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it's dangerous, so why are you doing it anyway? – akkkk Jun 24 '12 at 21:47
It is likely what your book expects you to do. The answer of experimentX shows you where the danger lurks. – ncmathsadist Jun 24 '12 at 21:49
Since the context is likely the real domain, this danger is suppressed. – ncmathsadist Jun 24 '12 at 21:50
Here is another pernicious peril. Notice that $(-1)^{1/3} = (-1)^{2/6} = (1)^{1/6} = 1.$ Oops. – ncmathsadist Jun 25 '12 at 1:21
you clearly missed my point. if it's so dangerous, at the very least you should mathematically explain why you can do it anyway. saying it's dangerous is easy - now explain why you can do it anyway. – akkkk Jun 25 '12 at 14:11

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