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 $(a,b)$ is equal to 

$S=\left(1+ax+b{x}^2\right)(1-2x{)}^{18}$

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

Coefficient of $x^3$ in the expansion of S is ${ }^{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

$\frac{544}{3}-17a+b=0$               (1)

Similarly, coefficient of $x^4$is

${ }^{18}{C}_4(-2{)}^4+a\left({ }^{18}{C}_3(-2{)}^3\right){+}^{18}{C}_2(-2{)}^{2}b=0$

Divide by ${ }^{18}{C}_2(-2{)}^2$ to obtain

$80-\frac{32}{3}a+b=0$                 (2)

Subtract (2) from (1) to obtain

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

From $b=17\times 16-\frac{544}{3}=\frac{272}{3}$