Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
1) Because of the negative exponent, flip, we want $(-243)^{2/3}$. 2) We will take a cube root, then square it. That will kill the minus sign. –  André Nicolas Jun 24 '12 at 21:32
    
@André: It depends on what you think $\sqrt[3]{-1}$ should be: your comment supposes that it is $-1$, but there are other approaches –  Henry Jun 24 '12 at 23:29
    
@Henry: In a problem tagged homework, arithmetic, that seems like the only available choice. –  André Nicolas Jun 24 '12 at 23:40

3 Answers 3

up vote 2 down vote accepted

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

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

share|improve this answer
1  
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

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

share|improve this answer

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

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.