Proof that in a triangle ABC

Question

a³sin(B-C)+b³sin(C-A)+c³sin(A-B)=0

Answer 1

prove: a³sin(B-C)+b³sin(C-A)+c³sin(A-B)=0

If you make an isosceles 45:45:90 degree triangle

where a=c, Angle A = 45 = Angle C and Angle B = 90 degrees,

a^3Sin(90-45) + b^3Sin(45-45) + c^3Sin(-45)

where a = c

a^3Sin(45) + b^3Sin(0) + a^3Sin(-45) = a^3Sin(45) - a^3Sin(45) = 0