Question : The value of expression $4\left(\sin ^6 A+\cos ^6 A\right)-6\left(\sin ^4 A+\cos ^4 A\right)+8$ is:
Option 1: 4
Option 2: 8
Option 3: 7
Option 4: 6
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Correct Answer: 6
Solution :
We know that $\left(\sin ^6 \theta+\cos ^6 \theta\right)=(\sin^2 \theta +\cos^2 \theta)(\sin^4 \theta + \cos^4 \theta-\sin^2 \theta\cos^2 \theta)$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)=4(1)(\sin^4 \theta + \cos^4 \theta-\sin^2 \theta\cos^2 \theta)$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)-6\left(\sin ^4 A+\cos ^4 A\right)=4(\sin^4 \theta + \cos^4 \theta)-4\sin^2 \theta\cos^2 \theta)-6\left(\sin ^4 A+\cos ^4 A\right)$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)-6\left(\sin ^4 A+\cos ^4 A\right)=-2(\sin^4 \theta + \cos^4 \theta+2\sin^2 \theta\cos^2 \theta)$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)-6\left(\sin ^4 A+\cos ^4 A\right)=-2(\sin^2 \theta +\cos^2 \theta)^2$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)-6\left(\sin ^4 A+\cos ^4 A\right)+8=-2(1)^2+8$
$⇒4\left(\sin ^6 \theta+\cos ^6 \theta\right)-6\left(\sin ^4 A+\cos ^4 A\right)+8=6$
Hence, the correct answer is 6.
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