64*Sin8x*(Sin^3(4x)+Cos^3(4x))Если Sin(4x)+Cos(4x)=1/2Если не затруднит подробно)

Решите задачу:

$(sin4x+cos4x)^2=(sin^24x+cos^24x)+2sin4x\cdot cos4x=\\\\=1+sin8x=(\frac{1}{2})^2\; \; \to \; \; sin8x=\frac{1}{4}-1=-\frac{3}{4}\\\\\\(sin4x+cos4x)^3=sin^34x+cos^34x+3sin^24x\cdot cos4x+\\\\+3sin4x\cdot cos^24x=sin^34x+cos^34x+3sin4x\cdot cos4x(sin4x+cos4x)=\\\\=(sin^34x+cos^34x)+\frac{3}{2}sin8x\cdot \frac{1}{2}=(sin^34x+cos^34x)+\frac{3}{4}\cdot \frac{-3}{4}\\\\\\(\frac{1}{2})^3=(sin^34x+cos^34x)-\frac{9}{16}\\\\sin^34x+cos^34x=\frac{1}{8}+\frac{9}{16}=\frac{11}{16}$

$64sin8x(sin^34x+cos^34x)=64\cdot \frac{-3}{4}\cdot \frac{11}{16}=-33$