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Решите задачу:

$8cos^{6}x+8sin^{6}x-3cos4x=$
$=8(cos^{4}x-cos^{2}xsin^{2}x+sin^{4}x)-3(cos^{2}2x-sin^{2}2x )=$
$=8(cos^{4}x-cos^{2}xsin^{2}x+sin^{4}x)-3((cos^{2}x-sin^{2}x)^{2}-4sin^{2}xcos^{2}x)=$
$=8cos^{4}x-8cos^{2}xsin^{2}x+8sin^{4}x-3(cos^{4}x+sin^{4}x-2sin^{2}xcos^{2}x-4sin^{2}xcos^{2}x)=$
$=8cos^{4}x-8cos^{2}xsin^{2}x+8sin^{4}x-3cos^{4}x+3sin^{4}x+18sin^{2}xcos^{2}x=$
$=5cos^{4}x+5cos^{2}xsin^{2}x+10sin^{4}x=5(cos^{2} x+sin^{2} x)^{2} =5$