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The top and bottom of the fraction both contain negative exponents. Since $c^{-3}$ on the bottom has a negative exponent, it is moved to the top of the fraction (numerator). Since the $d^{-3}$ on the top of the fraction has a negative exponent, it is moved to the bottom of the fraction (denominator). $$\frac{d^{-3}} {c^{-3}}$$ to $$\frac{c^3} {d^3}$$ Any explanation why does this happen?

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  • $\begingroup$ It comes from the rule $a^{b-c} = \frac{a^b}{a^c}$ Applying that rule $a^{-3} = a^{0-3} = \frac{a^0}{a^3} = \frac{1}{a^3}$ $\endgroup$ Nov 14, 2013 at 4:51
  • $\begingroup$ Is the edit giving what you intended. If so, you could at least use the ^ to indicate powers and parentheses to provide grouping. Even better, write it in $\LaTeX$. An intro is here $\endgroup$ Nov 14, 2013 at 4:54

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One of the laws of exponents is $a^{-b}=\frac 1{a^b}$. So the $d^{-3}$ in the numerator is equal to $\frac 1{d^3}$ and $\frac 1{c^{-3}}=c^3$. Putting them together gives what you are asking about.

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  • $\begingroup$ so typically what will happen is something like this? 1/d^3 / 1/c^3 ? $\endgroup$
    – Dave
    Nov 14, 2013 at 4:53
  • $\begingroup$ Yes, if you mean $\frac {d^{-3}}{c^{-3}}=\dfrac {\frac 1{d^3}}{\frac 1{c^3}}=\frac {c^3}{d^3}$. Please use parentheses when you use slashes. One can guess where the main division should be, but we shouldn't need to guess. $\endgroup$ Nov 14, 2013 at 4:56
  • $\begingroup$ Yup. that one. sorry, I am new here I don't here the shortcuts. from here 1/d^3 (/) 1/c^3 how did it came to c^3/d^3 $\endgroup$
    – Dave
    Nov 14, 2013 at 5:00
  • $\begingroup$ @Dave: You multiply numerator and denominator by $c^3d^3$ It is the usual technique to clear four level fractions like this-multiply by the product of the denominators. $\endgroup$ Nov 14, 2013 at 5:04
  • $\begingroup$ Finally got it. Thanks. $\endgroup$
    – Dave
    Nov 14, 2013 at 5:05

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