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I need to simplify this: $$\left(\dfrac{y^2}{-3x^3}\right)^3$$

I was able to simplify it to: $\dfrac{y^6}{-27x^9}$.

I am not sure if I need to move the negative sign outside of the fraction then, so that it reads negative in total. The negative sign in the denominator is what I am struggling with, how can I remove it?

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    $\begingroup$ Yes, you can write it outside the fraction, or even in the numerator. $\endgroup$ – M. Vinay Jun 5 '14 at 6:37
  • $\begingroup$ Yes, that comes from a proposition of fields. $\endgroup$ – jdoicj Jun 5 '14 at 6:42
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    $\begingroup$ @boywholived Overkill :P $\endgroup$ – M. Vinay Jun 5 '14 at 6:46
  • $\begingroup$ @M.Vinay: If you can show an easier proof, I will readily agree to your statement. $\endgroup$ – jdoicj Jun 5 '14 at 6:47
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    $\begingroup$ @boywholived Overkill given the OP's level, that's all I meant. $\endgroup$ – M. Vinay Jun 5 '14 at 6:50
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This may help you:

$$-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}$$

Considering the above, we can see that

$$ -\frac{y^6}{27x^9} = \frac{-y^6}{27x^9} = \frac{y^6}{-27x^9}$$

Although having the negative sign in the denominator is usually frowned upon.

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First of all, the power of the x on the denominator should be a 3 not a 9. Remember that one of the rules for exponents is: ${(x^m)}^n = x^{mn}$. That aside, I think the negative signs are usually written on the numerator in final answers.

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  • $\begingroup$ Actually, in the first post it was "x^", which I edited to $x^3$ (after seeing the $x^9$ in the next step). Looks like this was lost in the later edits. $\endgroup$ – M. Vinay Jun 5 '14 at 7:29

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