Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 16

$\frac{1}{16}(5+7cos2x+3cos4x+cos4x\cdot cos2x)$

Use $cos^2x=\frac{1+cos2x}{2}$, we have $cos^6x=(\frac{1+cos2x}{2})^3=\frac{1}{8}(1+3cos2x+3cos^22x+cos^32x) =\frac{1}{8}(1+3cos2x+3\times\frac{1+cos4x}{2}+\frac{1+cos4x}{2}\times cos2x) =\frac{1}{16}(5+7cos2x+3cos4x+cos4x\cdot cos2x)$