The ratio of the maximum value and the minimum value of $$\sin^6 x + \cos^6 x$$ is

Let y $$=\sin^6 x + \cos^6 x$$

y $$=(\sin^2 x)^3 + (\cos^2 x)^3$$

y $$=(\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)$$

y $$=1 - 3\sin^2 x \cos^2 x$$

y $$=1 - 3(\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(2\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(sin2 x)^2$$

When sin2x is maximum, y will be minumum and maximum value of sin2x is 1. 

y (min) = $$1 - \frac{3}{4} * 1 = \frac{1}{4}$$

When sin2x is minimum, y will be maximum and minimum value of sin2x is 0.

y (max) = $$1 - \frac{3}{4} * 0 = 1$$

Required ratio = 1: $$\frac{1}{4}$$ = 4: 1