Let $x_1, x_2, x_3, x_4$ are the roots of the equation

$x^4-4x^3+a{x}^2+bx+1=0$

$\therefore x_1+x_2+x_3+x_4=4$ and $x_1{x}_2{x}_3{x}_4=1$

$\because \mathrm{AM}=\frac{x_1+x_2+x_3+x_4}{4}=\frac{4}{4}=1$

and $GM={\left(x_1{x}_2{x}_3{x}_4\right)}^{1/4}=(1{)}^{1/4}=1$

i.e., $AM=GM$

which is true only when $x_1=x_2=x_3=x_4=1$

Hence, given equation has all roots identical, equal to $1$ i.e., equation have form

$(x-1{)}^4=0$

$\Rightarrow x^4-4x^3+6x^2-4x+1=0$

On comparing Eqs. (i) and (ii), we get

$a=6, b=-4$