Question

If ${(1+ax+b{x}^2)}^4=a_0+a_{1}x+a_2{x}^2+.........+a_8{x}^8$, where $a,b,a_0,a_1,.......a_8\in R$ such that $a_0+a_1+a_2\ne 0$ and $\left|\begin{array}{ccc}{a}_0& a_1& a_2\\ a_1& a_2& a_0\\ a_2& a_0& a_1\end{array}\right|=0$, then the value of $\frac{5a}{b}$ is ............

Answer 1

${(1+ax+b{x}^2)}^4=a_0+a_{1}x+a_2{x}^2+.........+a_8{x}^8$ ----(1)
such that $a_0+a_1+a_2\ne 0$
$\left|\begin{array}{ccc}{a}_0& a_1& a_2\\ a_1& a_2& a_0\\ a_2& a_0& a_1\end{array}\right|=0$
$\Rightarrow -(a_0+a_1+a_2)(a_0^2+a_1^2+a_2^2-a_0{a}_1-a_1{a}_2-a_2{a}_0)=0$
$\Rightarrow -\frac{1}{2}(a_0+a_1+a_2)\left(\right(a_0-a_1{)}^2+(a_1-a_2{)}^2+(a_2-a_0{)}^2)=0$
$\therefore a_0=a_1=a_2$ ($\because a_0+a_1+a_2\ne 0$)
put $x=0$ in (1)
$\Rightarrow a_0=1$
differentiating (1) on both sides gives
$4(1+ax+b{x}^2{)}^3(a+2bx)=a_1+2a_{2}x+\cdots +8a_8{x}^7$ -----(2)
put $x=0$ in (2)
$\Rightarrow 4a=a_1$
differentiating (2) on both sides gives
$4\left(\right(1+ax+b{x}^2{)}^3\left(2b\right)+3(a+2bx{)}^2(1+ax+b{x}^2{)}^2)=2a_2+6a_{3}x+\cdots +56a_8{x}^6$ ----(3)
put $x=0$ in (3)
$4(2b+3a^2)=2a_2$
$\therefore 1=4a=4b+6a^2$
$\Rightarrow a=\frac{1}{4},b=\frac{5}{32}$
$\therefore 5\frac{a}{b}=8$