The roots of the equation $x^3-14 x^2+56 x-64=0$ are in
The roots of the equation $x^3-14 x^2+56 x-64=0$ are in
AGP
$\mathrm{HP}$
$\mathrm{AP}$
GP
Solution
We have,
$\begin{aligned} & x^3-14 x^2+56 x-64=0 \\ \text { at } x & =2,8-14.4+56.2-64 \\ = & 8-56+112-64=0\end{aligned}$
Ther efore, the given equation can be written as
$\begin{array}{r}x^2(x-2)-12 x(x-2)+32(x-2)=0 \\ (x-2)\left(x^2-12 x+32\right)=0 \\ (x-2)(x-4)(x-8)=0\end{array}$
The roots are $2,4,8$, which are in GP.