If $(1-\mathrm{p})\left(1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5\right)=1-{\mathrm{p}}^6, \mathrm{p}\ne 1$, then the value of $\frac{\mathrm{p}}{\mathrm{x}}$ is

$(1-\mathrm{p})\left(1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5\right)=1-{\mathrm{p}}^6$

$\begin{array}{l}\mathrm{or} 1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5=\frac{1-{\mathrm{p}}^6}{1-\mathrm{p}}\\ \mathrm{or} 1+3\mathrm{x}+9{\mathrm{x}}^2+27{\mathrm{x}}^3+81{\mathrm{x}}^4+243{\mathrm{x}}^5=1+\mathrm{p}+{\mathrm{p}}^2+{\mathrm{p}}^3+{\mathrm{p}}^4+{\mathrm{p}}^5\end{array}$

Comparing, we get p = 3x or p/x = 3.