If 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 zeros, then is$(a, b)$ equal to

$\begin{array}{l}S=\left(1+ax+b{x}^2\right)(1-2x{)}^{18}\\ =\left(1+ax+b{x}^2\right)\left[1{+}^{18}{C}_1(-2x)+\left.{ }^{18}{C}_2(-2x{)}^2{+}^{18}{C}_3(-2x{)}^3{+}^{18}{C}_4(-2x{)}^4+\dots .\right]\right]\\ \end{array}$

Coefficient of  in $x^3$ the expansion of ${ }^{18}{C}_3(-2{)}^3+a\left({ }^{18}{C}_2(-2{)}^2\right){+}^{18}{C}_1(-2)b=0$

Divide by ${ }^{18}{C}_1(-2)$ to obtain  

$\frac{304}{3}-\frac{19}{3}a=0\Rightarrow a=16$

From $\left(1\right),b=17\times 16-\frac{544}{3}=\frac{272}{3}$