Let the term independent of $x$in the expansion of ${\left({\mathrm{x}}^2-\frac{1}{\mathrm{x}}\right)}^9$ has the value $p$ and $q$ be the sum of the coefficients of its middle terms, then $\left(p-q\right)$  equals

$$\begin{gathered} {\mathrm{T}}_{\mathrm{r}+1}\text{\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^2-\frac{{\displaystyle 1}}{{\displaystyle \mathrm{x}}}\right)}^9\text{is\hspace{0.17em}\hspace{0.17em}}{}^9\mathrm{C}_{\mathrm{r}}{\mathrm{x}}^{2(9-\mathrm{r})}{\left(-\frac{{\displaystyle 1}}{{\displaystyle \mathrm{x}}}\right)}^{\mathrm{r}} \\ {}^9\mathrm{C}_{\mathrm{r}}.{\mathrm{x}}^{18-3\mathrm{r}}.{(-1)}^{\mathrm{r}} \end{gathered}$$
for term independent of x, $18-3\mathrm{r}=0\Rightarrow \mathrm{r}=6$
$\therefore$7th term is independent of x and equals ${}^9\mathrm{C}_6={}^9\mathrm{C}_3=84$
Also there are 10 terms, hence 5th term and 6th are the two middle term 
$$\begin{gathered} {\mathrm{T}}_5={}^9\mathrm{C}_4.{\mathrm{x}}^6 \\ {\mathrm{T}}_6=-{}^9\mathrm{C}_5.{\mathrm{x}}^3 \end{gathered}$$
$\therefore$q = coefficient of 5^th + coefficient of 6th term
$$\begin{gathered} {}^9\mathrm{C}_4={}^9\mathrm{C}_5=0 \\ \mathrm{hence} \mathrm{p}=84;\mathrm{q}=0 \\ \mathrm{p}-\mathrm{q}={}^9\mathrm{C}_3 \end{gathered}$$