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

For a cubic equation x^3 - ax^2 + bx - c = 0, if roots are in GP (a/r, a, ar), then product of roots = c = a^3. Here 64 = a^3, so a = 4. If 4 is a root, 64 - 14(16) + 56(4) - 64 = 64 - 224 + 224 - 64 = 0. Yes, 4 is a root. The other roots are 4/r and 4r. Sum of roots = 4/r + 4 + 4r = 14 => 4/r + 4r = 10 => 2/r + 2r = 5 => 2r^2 - 5r + 2 = 0 => (2r-1)(r-2)=0. r=2. Roots are 2, 4, 8. They are in GP.