If $\mathrm{p}>0, \mathrm{x}>0, \mathrm{p}\ne 1\text{ and }(1-\mathrm{p})\left(1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5\right)=\left(1-{\mathrm{p}}^6\right)\text{ then }\frac{\mathrm{p}}{\mathrm{x}}=$

$\begin{array}{l}\mathrm{p}>0, \mathrm{x}>0, \mathrm{p}\ne 1\\ (1-\mathrm{p})\left(1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5\right)\\ \Rightarrow 1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5=\frac{1-{\mathrm{p}}^6}{1-\mathrm{p}}\\ 1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5=1+\mathrm{p}+{\mathrm{p}}^2+p^3+p^4+p^5\\ p^5=243{\mathrm{x}}^5\Rightarrow \mathrm{p}=3\mathrm{x}\therefore \frac{\mathrm{p}}{\mathrm{x}}=3\end{array}$