If the coefficient of $x^2 and x^3$ are both zero, in the expansion of the expression $(1+ax+b{x}^2){(1–3x)}^{15},$ in powers of x, then the ordered pair (a,b) is equal to

 $\begin{array}{rcl} & & (1+\mathrm{a} \mathrm{x}+\mathrm{b} {\mathrm{x}}^2) {(1-3\mathrm{x})}^{15}\\ & & (1+\mathrm{a} \mathrm{x}+\mathrm{b} {\mathrm{x}}^2) \left(1-{}^{15}{\mathrm{C}}_1\left(3\mathrm{\theta }\right)+{}^{15}{\mathrm{C}}_2{\left(3\mathrm{\theta }\right)}^2-{}^{15}{\mathrm{C}}_3{\left(3\mathrm{\theta }\right)}^3\right)\end{array}$

Coeff of ${\mathrm{x}}^2={}^{15}{\mathrm{C}}_2 3^2-\mathrm{a} 15\left(3\right)+\mathrm{b}=0$

               $\Rightarrow 945-45\mathrm{a}+\mathrm{b}=0 -\left(1\right)$

Coeff ${\mathrm{x}}^3={}^{-15}{\mathrm{C}}_3 {\left(3\right)}^3+{}^{15}{\mathrm{C}}_2 \mathrm{a}{\left(3\right)}^2+{}^{15}{\mathrm{C}}_2 \mathrm{a}{\left(3\right)}^2-{}^{15}{\mathrm{C}}_1 3\mathrm{b}=0$

               $\Rightarrow -12, 285+945 \mathrm{a}-45 \mathrm{b}=0 -\left(2\right)$

By Solving (1) & (2)  $\mathrm{a}=28, \mathrm{b}=315$