If x is small so that  $x^2$ and higher powers can be neglected, then the approximate  value of   $\frac{{\left(1-2x\right)}^{-1}.{\left(1-3x\right)}^{-2}}{{\left(1-4x\right)}^{-3}}$  is

By neglecting  $x^2$  and higher power 

${\left(1-2x\right)}^{-1}=1+2x$

${\left(1-3x\right)}^{-2}=1+6x,{\left(1-4x\right)}^{-3}=1+12x$

$\Rightarrow \frac{{\left(1-2x\right)}^{-1}{\left(1-3x\right)}^{-2}}{{\left(1-4x\right)}^{-3}}$

$=\frac{\left(1+2x\right)\left(1+6x\right)}{\left(1+12x\right)}=\left(1+8x\right)\left(1-12x\right)$

$=1+8x-12x-96x^2=1-4x$