Question

Prove that sin^6 thetha+cos^6 thetha=3 sin^4 thetha+3cos^2 thetha-2
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Answer 1

${\huge{\fbox{\red{answer}}}}$

$\tt{sin^6\:\theta+cos^6\:\theta+3 sin^2\: \theta\:cos^2=1} \\\\\implies \tt{LHS =(sin^2\:\theta)^3\:(cos^2\:\theta)^3 +3sin^2\:\theta \:cos^2 \:\theta}\\ \\ \tt{Using, \:[a^3+b^3=(a+b)^3+3ab(a+b)]} \\\\ \implies \tt{LHS=(sin^2\:\theta\:+cos^2\:\theta)^3-3sin^2\:\theta\:cos^2\:\theta\:(sin^2\:\theta\:+cos^2)^3+3sin^2+\:\theta\:cos^2\:\theta}\\\\\implies\tt{LHS=1-3sin^2\:\theta\:cos^2\:\theta\:+3sin^2\:\theta\:cos^2\:\theta=1=RHS}$

HOPE IT HELPED! :)

Answer 2

Replace alpha with theta

LHS=RHS

Hence proved