The term independent of x in the expansion of $(1+ax+b{x}^2){(1-2x)}^{18}$ in powers of x are both zero, then $(a, b) =$     

$(1+ax+b{x}^2){(1-2x)}^{18}={(1-2x)}^{18}+ax{(1-2x)}^{18}+b{x}^2{(1-2x)}^{18}$
$\begin{array}{l}Writecoefficientof{x}^3=0\&coefficientof{x}^4=0\\ Weget153a-9b=1632-->\left(1\right)\\ 3b-32a=-240-->\left(2\right)\\ Solvea=16b=\frac{272}{3}.\end{array}$