Two problems with Exponents

Question

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!

Answer 1

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 $$\begin{align} & \frac1{1+p^{a-b}+p^{a-c}}+\frac1{1+p^{b-c}+p^{b-a}}+\frac1{1+p^{c-a}+p^{c-b}}\\ & =\frac{p^{-a}}{p^{-a}+p^{-b}+p^{-c}}+\frac{p^{-b}}{p^{-b}+p^{-c}+p^{-a}}+\frac{p^{-c}}{p^{-c}+p^{-a}+p^{-b}}\\ & =\frac{p^{-a}+p^{-b}+p^{-c}}{p^{-a}+p^{-b}+p^{-c}}\\ & =1\end{align}$$