# Two problems with Exponents

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!

• I have edited it. Please oblige. Apr 11, 2016 at 13:23
• 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 Apr 11, 2016 at 13:45

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}
• @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. Apr 11, 2016 at 14:35
• @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. Apr 11, 2016 at 19:58