The term independent of x in the expansion of ${(1-x)}^2.{\left(x+\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{10}$  is

${(1-x)}^2.{\left(x+\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{10}$

consider

${\left(x+\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^{10}$

$T_{r+1}={}^{10}{\text{C}}_r{x}^{10-r}{\left(\frac{{\displaystyle 1}}{{\displaystyle x}}\right)}^r$
$\begin{array}{l}={}^{10}{\text{C}}_r{x}^{10-2r}\\ 10-2r=0\\ r=5\end{array}$

$\therefore$ const term$={}^{10}{\text{C}}_5$

For $\frac{1}{x}$ term $10-2r=-1$

$r=\frac{11}{2}\left(\text{not\hspace{0.17em}\hspace{0.17em}possible}\right)$

For $\frac{1}{x^2}$ term $10-2r=-2$

$\begin{array}{l}12=2r\\ r=6\\ \therefore \frac{1}{x^2}\text{coefficient}={}^{10}{\text{C}}_6\end{array}$

$\therefore$ Independent term $=1\times {}^{10}{\text{C}}_5+1\times {}^{10}{\text{C}}_6$

$={}^{10}{\text{C}}_5+{}^{10}{\text{C}}_6$

$={}^{11}{\text{C}}_6[\because {}^n{\text{C}}_r+{}^n{\text{C}}_{r-1}={}^{n+1}{\text{C}}_r]$

$={}^{11}{\text{C}}_5$