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$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}$
$+4\left(\sin ^{6} x+\cos ^{6} x\right)$
is equal to
Options:
$+4\left(\sin ^{6} x+\cos ^{6} x\right)$
is equal to
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The correct answer is:
13
$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}$
$+4\left(\sin ^{6} x+\cos ^{6} x\right)$
$=3(1-2 \sin x \cos x)^{2}+6(1+2 \sin x \cos x)$
$\quad+4\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x\right.$
$\left.\quad-\sin ^{2} x \cos ^{2} x\right)$
$=3\left[1+4 \sin ^{2} x \cos ^{2} x-4 \sin x \cos x\right]$
$\quad+6+12 \sin x \cos x+4\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}\right.$
$\left.\quad-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x\right]$
$+4\left(\sin ^{6} x+\cos ^{6} x\right)$
$=3(1-2 \sin x \cos x)^{2}+6(1+2 \sin x \cos x)$
$\quad+4\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x\right.$
$\left.\quad-\sin ^{2} x \cos ^{2} x\right)$
$=3\left[1+4 \sin ^{2} x \cos ^{2} x-4 \sin x \cos x\right]$
$\quad+6+12 \sin x \cos x+4\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}\right.$
$\left.\quad-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x\right]$
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