If in a triangle ABC, sin³A + sin³B + sin³C = 3 sinA sinB sinC, then what is the value of the determinant | a b c ; b c a ; c a b |, where a, b, c are sides of the triangle?

The identity sin³A + sin³B + sin³C = 3sinA sinB sinC implies sinA + sinB + sinC = 0 (using x³+y³+z³ – 3xyz = (x+y+z)(x²+y²+z²–xy–yz–zx)), but since sines of angles of a triangle are positive, the only way is if a + b + c = 0 (by sine rule, proportionality). The determinant | a b c ; b c a ; c a b | expanded equals –(a³ + b³ + c³ – 3abc) = –(a+b+c)(a²+b²+c²–ab–bc–ca). If a + b + c = 0, the determinant equals 0.