If ${\mathrm{p}}^4+{\mathrm{q}}^3=2(\mathrm{p}>0, \mathrm{q}>0)$ then the maximum value of term independent of $x$ in the expansion of ${\left({\mathrm{px}}^{\frac{1}{12}}+{\mathrm{qx}}^{-\frac{1}{9}}\right)}^{14}$ is

We know that in the expansion of ${\left(a{x}^p+\frac{b}{x^q}\right)}^n$ , if the term independent of $x$ is $T_{r+1}$ then $r=\frac{np}{p+q}$

Hence, the term indpendent of $x$ in the expansion of ${\left({\mathrm{px}}^{\frac{1}{12}}+{\mathrm{qx}}^{-\frac{1}{9}}\right)}^{14}$

Hence, 

          $\begin{array}{rcl}\mathrm{r}& =& \frac{\mathrm{np}}{\mathrm{p}+\mathrm{q}}\\ & =& \frac{14·{\displaystyle \frac{1}{12}}}{{\displaystyle \frac{1}{12}}+{\displaystyle \frac{1}{9}}}\\ & =& 6\end{array}$

Hence, the term independent of $x$ is ${}^{14}C_6{p}^8{q}^6$

Since the ${\mathrm{p}}^4+{\mathrm{q}}^3=2$ the maximum value of $p^8{q}^6$ is 1, since AM is greater than GM

Hence, the maximum value of the term independent of $x$ is $\boxed{{}^{14}C_6}$