Question
Prove that: $\sin ^6 \theta+\cos ^6 \theta=1-3 \sin ^2 \theta \cos ^2 \theta$.
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Answer
$\sin ^6 \theta+\cos ^6 \theta$
$=\left(\sin ^2 \theta\right)^3+\left(\cos ^2 \theta\right)^3$
$=\left(\sin ^2 \theta+\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^2 \theta-\sin ^2 \theta \cos ^2 \theta\right)$
$=1 \times\left(\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2+2 \sin ^2 \theta \cdot \cos ^2 \theta-3 \sin ^2 \theta \cdot \cos ^2 \theta\right)$
$=\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2-3 \sin ^2 \theta \cdot \cos ^2 \theta$
$=1-3 \sin ^2 \theta \cos ^2 \theta$
$=\text { RHS }$
$=\left(\sin ^2 \theta\right)^3+\left(\cos ^2 \theta\right)^3$
$=\left(\sin ^2 \theta+\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^2 \theta-\sin ^2 \theta \cos ^2 \theta\right)$
$=1 \times\left(\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2+2 \sin ^2 \theta \cdot \cos ^2 \theta-3 \sin ^2 \theta \cdot \cos ^2 \theta\right)$
$=\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2-3 \sin ^2 \theta \cdot \cos ^2 \theta$
$=1-3 \sin ^2 \theta \cos ^2 \theta$
$=\text { RHS }$
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