Prove that 2 ( ^6 A + ^6 A )-3 ( ^4 A + ^4 A )+1=0. - Vidyadip

Question

Prove that $2\left(\sin ^6 A +\cos ^6 A \right)-3\left(\sin ^4 A +\cos ^4 A \right)+1=0$.

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Answer

$\sin ^6 A+\cos ^6 A=\left(\sin ^2 A\right)^3+\left(\cos ^2 A\right)^3$
$=\left(1-\cos ^2 A\right)^3+\left(\cos ^2 A\right)^3 \quad \ldots \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =1-3 \cos ^2 A+3\left(\cos ^2 A\right)^2-\left(\cos ^2 A\right)^3+\cos ^6 A$
$ =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$
$ \sin ^4 A+\cos ^4 A=\left(\sin ^2 A\right)^2+\left(\cos ^2 A\right)^2$
$ =\left(1-\cos ^2 A\right)^2+\left(\cos ^2 A\right)^2$
$ =1-2 \cos ^2 A+\left(\cos ^2 A\right)^2+\left(\cos ^2 A\right)^2 \ldots \ldots\left[\because(a-b)^2=a^2-2 a b+b^2\right]$
$ =1-2 \cos ^2 A+2 \cos ^4 A$
$ =1-2 \cos ^2 A\left(1-\cos ^2 A\right)$
$ =1-2 \cos ^2 A \sin ^2 A$
$ \text { L.H.S }=2\left(\sin ^6 A+\cos ^6 A\right)-3\left(\sin ^4 A+\cos ^4 A\right)+1$
$ =2\left(1-3 \cos ^2 A \sin ^2 A\right)-3\left(1-2 \cos ^2 A \sin ^2 A\right)+1$
$ =2-6 \cos ^2 A \sin ^2 A-3+6 \cos ^2 A \sin ^2 A+1$
$ =0$
$ =\text{R.H.S}$
$ \therefore 2\left(\sin ^6 A+\cos ^6 A\right)-3\left(\sin ^4 A+\cos ^4 A\right)+1=0$

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