Mathematics - Binomial Theorem Question with Solution | TestHub

$x=0:a_0=1$ Differentiations $4{\left(1+ax+b{x}^2\right)}^3(a+2bx)=a_1+2a_{2}x+\dots$. Let $x=04a=a_1$ The value of the determinant is $-\left(a_0^3+a_1^3+a_2^3-3a_0{a}_1{a}_2\right)=0$ $\Rightarrow \because a_0+a_1+a_2\ne 0$ the $a_0=a_1=a_2=4a$ Differentially again and putting $x=0$ $a=\frac{1}{4}:b=\frac{5}{32}$