Помогите, пожалуйста, с решением

$sin2\alpha=2sin\alpha cos\alpha\ /()^2$

$sin^22\alpha=4sin^2\alpha cos^2\alpha\ /\cdot(-2)$

$-2sin^22\alpha=-8sin^2\alpha cos^2\alpha\ /+1$

$1-2sin^22\alpha=1-8sin^2\alpha cos^2\alpha$

$cos4\alpha=1-8sin^2\alpha cos^2\alpha$

$8sin^2\alpha cos^2\alpha=1-cos4\alpha\ /:8$

$sin^2\alpha cos^2\alpha= \frac{1}{8} - \frac{1}{8} cos4\alpha$

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$sin^6\alpha+cos^6\alpha=$

$(sin^2\alpha)^3+(cos^2\alpha)^3=$

$(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)=$

$sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha=$

$(sin^4\alpha+cos^4\alpha)-sin^2\alpha cos^2\alpha=$

$(sin^2\alpha+cos^2\alpha)^2-2sin^2\alpha cos^2\alpha-sin^2\alpha cos^2\alpha=$

$1-3sin^2\alpha cos^2\alpha=$

$1-3 \cdot \left( \frac{1}{8} - \frac{1}{8} cos4\alpha \right) =$

$1- \frac{3}{8}+ \frac{3}{8} cos4\alpha =$

$\frac{5}{8}+ \frac{3}{8} cos4\alpha$