If the coefficients of x^2 and x^3, are both zero, in the expansion of the expression (1+ax+bx^2)(1-3x)^15, in powers of x , then the ordered pair…

$$(1+ax+b{x}^{2}){(1-3x)}^{15}=(1+ax+b{x}^{2})(C015{-}^{15}{C}_{1} 3x {+}^{ 15}{C}_{2} {(3x)}^{2}{-}^{15}{C}_{3}{(3x)}^{3}+\ldots {-}^{15}{C}_{15}{(3x)}^{15})$$(1+ax+bx2)(1−3x)15=(1+ax+bx2)(C015−15C1​3x+15C2​(3x)2−15C3​(3x)3+…−15C15​(3x)15)

$\therefore$∴ Coefficient of ${x}^{2}$x2 is: ${ b.}^{15}{C}_{0}-3.{a}^{15}{C}_{1}+{3}^{2.}{.}^{15}{C}_{2}=0$b.15C0​−3.a15C1​+32..15C2​=0

(Given)

$$\Rightarrow b-45a+945=0......(i)$$⇒b−45a+945=0......(i)

Coefficient of ${x}^{3}$x3 is:

$$-b.3{.}^{15}{C}_{1}+9a{.}^{15}{C}_{2}-27{.}^{15}{C}_{3}=0$$−b.3.15C1​+9a.15C2​−27.15C3​=0

$$\Rightarrow -45b+9a\times \frac{15}{2}\times \frac{14}{1}-27\times \frac{15}{3}\times \frac{14}{2}\times \frac{13}{1}=0$$⇒−45b+9a×215​×114​−27×315​×214​×113​=0

$$\Rightarrow -b+21a-273=0$$⇒−b+21a−273=0

$$\Rightarrow b-21a+273=0.......(\mathrm{ii})$$⇒b−21a+273=0.......(ii)

From $(i)$(i) and $(\mathrm{ii})$(ii)

$a=28$a=28 and $b=315$b=315