The ratio of the coefficient of $x^{10}$ in ${(1-x^2)}^{10}$ and the term independent of $\text{'}x\text{'}$ in ${\left(x-\frac{{\displaystyle 2}}{{\displaystyle x}}\right)}^{10}$ is

Given, expansion is ${(1-x^2)}^{10}$

$\begin{array}{l}{T}_{r+1}={}^n{C}_r{\left(1\right)}^{n-r}{(-x^2)}^r\\ ={}^{10}{C}_r{\left(1\right)}^{10-r}{\left(x\right)}^{2r}{(-1)}^{2r}\end{array}$

for the $x^{10}$ let $2r=10\Rightarrow r=5$

$\therefore$ 6^th term $T_6=T_{5+1}={}^{10}{C}_5{\left(1\right)}^{10-5}{\left(x\right)}^{2\left(5\right)}{(-1)}^{2\left(5\right)}$

Coefficient of $x^{10}={}^{-10}{C}_5$

Given, expression is ${\left(x-\frac{{\displaystyle 2}}{{\displaystyle x}}\right)}^{10}$

$T_{r+1}={}^{10}{C}_r{\left(x\right)}^{10-r}{\left(\frac{{\displaystyle -2}}{{\displaystyle x}}\right)}^r$

$={}^{10}{C}_r{\left(x\right)}^{10-r}{(-2)}^5{\left(x\right)}^{-r}={}^{10}{C}_r{\left(x\right)}^{10-2r}{(-2)}^5$

for independent of $\text{'}x\text{'}$ i.e., index of $x=0$

$\begin{array}{l}10-2r=0\\ 2r=10\end{array}$

6^th term

$T_6=T_{5+1}={}^{10}{C}_5{\left(x\right)}^{10-2\left(5\right)}{(-2)}^5$

$=-{\left(2\right)}^5{}^{10}{C}_5$

The ratio is ${ }^{-10}{C}_5:-{\left(2\right)}^5{}^{10}{C}_5=\frac{+{}^{10}{C}_5}{+{\left(2\right)}^5{}^{10}{C}_5}$

$=1:2^5=1:32$.