If the Coefficient of  $x^3$ and $x^4$   in the expansion of

$\left.(1-ax+b{x}^2\right)(1-2x{)}^{18}$ in powers of $x$  are both zero then $(a, b)$  is 

equal to 

We have 

 $\begin{array}{l}\left(1+ax+b{x}^2\right)(1-2x{)}^{18}\\ =\left(1+ax+b{x}^2\right)\sum _{r=0}^{18}{ }^{18}{C}_r(-1{)}^r{2}^r{x}^r\\ =\sum _{r=0}^{18}{ }^{18}{C}_r(-1{)}^r{2}^r{x}^r+a\sum _{r=0}^{18}{ }^{18}{C}_r(-1{)}^r{2}^r{x}^{r+1}\end{array}$

                           $+b\sum _{r=0}^{18}{ }^{18}{C}_r(-1{)}^r{2}^r{x}^{r+2}$

The coefficient of   $x^3$ and $x^4$  in the expansion of  $\left(1+ax+b{x}^2\right)$

$(1-2x{)}^{18}$ are both zero. 

 ${\therefore }^{ 18}{C}_3(-1{)}^3{2}^3+a^{18}{C}_2(-1{)}^2{2}^2+b^{18}{C}_1(-1{)}^1{2}^1=0$

 and ${ }^8{C}_4(-1{)}^4{2}^4+a^{18}{C}_3(-1{)}^3{2}^3+b^{18}{C}_2(-1{)}^2{2}^2=0$

 $\Rightarrow 51a-3b=144$ and $32a-3b=240$ 

 $\Rightarrow a=16,b=\frac{272}{3}$