The roots of the equation ${\mathrm{x}}^3–14{\mathrm{x}}^2+56\mathrm{x}–64=0$ are in

$\begin{array}{l}GivenEquation{x}^3-14x^2+56x-64=0\\ Leta=1,b=-14,c=56,d=-64\\ IftherootsareinAPthena{b}^3+27a^{2}d=9abc\\ \Rightarrow 1{(-14)}^3+27\left(1\right)(-64)\ne 9\left(1\right)(-14)516\\ IftherootsareinGPthena{c}^3=b^{3}d\\ \therefore 1{\left(56\right)}^3={(-14)}^3(-64)\\ \Rightarrow 175616=175616\end{array}$