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Maximum and minimum values of sin⁶x+cos⁶x is

$\sin^{6} x + \cos^{6} x = (\sin^{2} x + \cos^{2} x)^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x) f (x) = 1 - 3 \sin^{2} x \cos^{2} x = 1 - \frac{3}{4} \sin^{2} 2 x - 1 \leq \sin 2 x \leq 1 M i n i m u m v a l u e = 1 - \frac{3}{4} (1) = \frac{1}{4} M a x i m u m v a l u e = 1$

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