In a triangle ABC,
a3cos(B−C)+b3cos(C−A)+c3cos(A−B)=
3 abc
a3cos(B−C)+b3cos(C−A)+c3cos(A−B)=k3sin3Acos(B−C)+k3sin3Bcos(C−A)+k3sin3Ccos(A−B)=12k3[sin2A(sin2B+sin2C)+sin2B(sin2C+sin2A)+sin2C(sin2A+sin2B)]=12k3[sin2A(2sinBcosB+2sinCcosC)+sin2B(2sinCcosC+2sinAcosA)+sin2C(2sinAcosA+2sinBcosB)]=k3[sinAsinB(sinAcosB+cosAsinB)+sinBsinC(sinBcosC+cosBsinC)+sinCsinA(sinCcosA+cosCsinA)]=k3[sinAsinB(sin(A+B))+sinBsinC(sin(B+C))+sinCsinA(sin(C+A))]=k3[sinAsinB(sin(π−C))+sinBsinC(sin(π−A))+sinCsinA(sin(π−B))]=k3[sinAsinBsinC+sinBsinCsinA+sinCsinAsinB]=3k3sinAsinBsinC=3abc.