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Question

In triangle ABC, let a,b,c be the lengths of the sides opposite to the angles A,B,C respectively. Let the value of a3cos3B+3a2bcos(A2B)+3ab2cos(2AB)+b3cos3A be l. Then the value of lc3 is equal to

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Solution

We have c=acosB+bcosA (By projection rule)
and asinBbsinA=0. (By sine rule)
Thus, we can write c=a(cosBisinB)+b(cosA+isinA) where i=1
c3=(aeiB+beiA)3c3=a3ei3B+3a2bei2BeiA+3ab2eiBei2A+b3ei3Ac3=l
(Consider real part only as in LHS, we have real value only)
lc3=1

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