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How to solve following problems on exponents:

$$\frac1{1+p^{a-b}+p^{a-c}}+\frac1{1+p^{b-c}+p^{b-a}}+\frac1{1+p^{c-a}+p^{c-b}}=?$$

and

If $a^2bc^2=5^3$ and $ab^2=5^6$, what is $abc$?

Please mention the method by which the result is derived!

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  • $\begingroup$ I have edited it. Please oblige. $\endgroup$
    – J_B892
    Apr 11, 2016 at 13:23
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    $\begingroup$ In the first, are $a,b,c$ required to be positive integers? If so, $a=c=1,b=5^3$. The second is simply 1/1=1 $\endgroup$
    – almagest
    Apr 11, 2016 at 13:45

1 Answer 1

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The first system of equations read $$\begin{align}a^2bc^2 & =5^3 &(1)\\ ab^2 & =5^6 & (2)\end{align}$$ If we square $(1)$ and divide by $(2)$, $$a^3c^4=\frac{\left(a^2bc^2\right)^2}{ab^2}=\frac{\left(5^3\right)^2}{5^6}=1$$ Squaring $(2)$ and dividing by $(1)$, $$\frac{b^3}{c^2}=\frac{\left(ab^2\right)^2}{a^2bc^2}=\frac{\left(5^6\right)^2}{5^3}=5^9$$ So we have $$a=c^{-\frac43}$$ $$b=125c^{\frac23}$$ Then $$abc=c^{-\frac43}\cdot125c^{\frac23}\cdot c=125\sqrt[3]c$$ The second equation is $$\begin{align} & \frac1{1+p^{a-b}+p^{a-c}}+\frac1{1+p^{b-c}+p^{b-a}}+\frac1{1+p^{c-a}+p^{c-b}}\\ & =\frac{p^{-a}}{p^{-a}+p^{-b}+p^{-c}}+\frac{p^{-b}}{p^{-b}+p^{-c}+p^{-a}}+\frac{p^{-c}}{p^{-c}+p^{-a}+p^{-b}}\\ & =\frac{p^{-a}+p^{-b}+p^{-c}}{p^{-a}+p^{-b}+p^{-c}}\\ & =1\end{align}$$

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  • $\begingroup$ I added labels to your equations to help with readability. Feel free to rollback if you'd prefer not to have them. $\endgroup$
    – zz20s
    Apr 11, 2016 at 14:06
  • $\begingroup$ In the second sum have you taken p to the power -a common? If so I can take even p to the power -b or p to the power -c common rite? $\endgroup$
    – J_B892
    Apr 11, 2016 at 14:21
  • $\begingroup$ @zz20s, thanks for the edits. In the second equation, I multiplied numerator and denominator of the first fraction by $p^{-a}$, numerator and denominator of the second by $p^{-b}$, and numerator and denominator of the third by $p^{-c}$. Exploiting the symmetry between the expressions in this way I was able to obtain a common denominator and so to obtain their sum. $\endgroup$ Apr 11, 2016 at 14:35
  • $\begingroup$ @zz20s Did you know that \tag 1 also produces $(1)$ to the right of the equation? It puts the label all the way at the right margin, which is both good (people expect to find it there) and bad (far from the actual equation in many cases--the text in Stackexchange is wider than it would be in many books). If the chosen format was a personal preference, OK, I just wasn't sure. $\endgroup$
    – David K
    Apr 11, 2016 at 19:58
  • $\begingroup$ I considered using \tag and decided against it, since, as you said, it was too far away from the equation. If you feel it would be better, by all means go ahead and change it. Ultimately, I just felt that having the equations labelled would just make the answer a touch more readable. $\endgroup$
    – zz20s
    Apr 11, 2016 at 20:19

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