If the coefficients of $x^{2}$ and $x^{3}$ are both zero, in the expansion of the expression $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ in powers of $\mathrm{x}$, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to :
Correct Option: , 3
Given expression is $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$
Co-efficient of $x^{2}=0$
$\Rightarrow{ }^{15} \mathrm{C}_{2}(-3)^{2}+a \cdot{ }^{15} \mathrm{C}_{1}(-3)+b \cdot{ }^{15} \mathrm{C}_{0}=0$
$\Rightarrow \frac{15 \times 14}{2} \times 9-15 \times 3 a+b=0$
$\Rightarrow 945-45 a+b=0$ ...........(i)
Now, co-efficient of $x^{3}=0$
$\Rightarrow{ }^{15} \mathrm{C}_{3}(-3)^{3}+a \cdot{ }^{15} \mathrm{C}_{2}(-3)^{2}+b \cdot{ }^{15} \mathrm{C}_{1}(-3)=0$
$\Rightarrow \frac{15 \times 14 \times 13}{3 \times 2} \times(-3 \times 3 \times 3)+a \times \frac{15 \times 14 \times 9}{2}$
$-b \times 3 \times 15=0$
$\Rightarrow 15 \times 3[-3 \times 7 \times 13+a \times 7 \times 3-b]=0$
$\Rightarrow 21 a-b=273$..(ii)
From (i) and (ii), we get,
$a=28, b=315 \Rightarrow(a, b) \equiv(28,31,5)$