Question
Prove that $\sin ^6 A+\cos ^6 A=1-3 \sin ^2 A \cdot \cos ^2 A$
✓
Answer
$ \text { L.H.S }=\sin ^6 A+\cos ^6 A$
$ =\left(\sin ^2 A\right)^3+\left(\cos ^2 A\right)^3$
$=\left(1-\cos ^2 A\right)^3+\left(\cos ^2 A\right)^3 \ldots \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =1-3 \cos ^2 A+3\left(\cos ^2 A\right)^2-\left(\cos ^2 A\right)^3+\cos ^6 A \ldots \ldots\left[\because(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right]$
$ =1-3 \cos ^2 A\left(1-\cos ^2 A\right)-\cos ^6 A+\cos ^6 A$
$ =1-3 \cos ^2 A \sin ^2 A$
$ =\text { R.H.S }$
$ \therefore \sin ^6 A+\cos ^6 A=1-3 \sin ^2 A \cdot \cos ^2 A$
$ =\left(\sin ^2 A\right)^3+\left(\cos ^2 A\right)^3$
$=\left(1-\cos ^2 A\right)^3+\left(\cos ^2 A\right)^3 \ldots \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =1-3 \cos ^2 A+3\left(\cos ^2 A\right)^2-\left(\cos ^2 A\right)^3+\cos ^6 A \ldots \ldots\left[\because(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right]$
$ =1-3 \cos ^2 A\left(1-\cos ^2 A\right)-\cos ^6 A+\cos ^6 A$
$ =1-3 \cos ^2 A \sin ^2 A$
$ =\text { R.H.S }$
$ \therefore \sin ^6 A+\cos ^6 A=1-3 \sin ^2 A \cdot \cos ^2 A$
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