In a triangle ABC, what is the value of
(a^3)*cos(3B)+3(a^2)b*cos(2B-A)+3a(b^2)*cos(B-2A)+(b^3)*cos(3A),
where a, b, c, A, B, C have their usual meanings?

Level: Senior

Let x = (a^3)*cos(3B)+3(a^2)b*cos(-2B+A)+3a(b^2)*cos(-B+2A)+(b^3)*cos(3A).
We have
x
= (a^3)*cos(3B)+3(a^2)b*cos(-2B+A)+3a(b^2)*cos(-B+2A)+(b^3)*cos(3A)
= Re{(a^3)[e^(-3Bi)]+3(a^2)b[e^(-2Bi+Ai)]+3a(b^2)[e^(-Bi+2Ai)]+(b^3)[e^(3Ai)]}
= Re{[ae^(-Bi)+be^(iA)]^3}.
The law of sines implies that a/sin(A) = b/sin(B), so a*sin(B)-b*sin(A) = 0. It is also well-known that a*cos(B)+b*cos(A) = c. Using these two facts,
x
= [a*cos(-B)+a*sin(-B)*i+b*cos(A)+b*sin(A)*i]^3
= [a*cos(B)-a*sin(B)*i+b*cos(A)+b*sin(A)*i]^3
= [a*cos(B)+b*cos(A)-i(a*sin(B)-b*sin(A))]^3^3
= [c-i*0]^3
= c^3.