The correct option is B 3sinAsinBsinC
Δ=∣∣
∣∣abcbcacab∣∣
∣∣=(a+b+c)∣∣
∣∣1bc1ca1ab∣∣
∣∣
[applying C1→C1+C2+C3 and taking (a+b+c) common from C1]=(a+b+c)∣∣
∣∣1bc0c−ba−c0a−bb−c∣∣
∣∣[applying R2→R2−R1 and R3→R3−R1]=(a+b+c)[(c−b)(b−c)−(a−b)(a−c)]=(a+b+c)[ab+bc+ca−a2−b2−c2]=−(a3+b3+c3)+3abc
Now, Δ=0 implies a3+b3+c3=3abc. Therefore, by the law of sines.
sinAa=sinBb=sinCc, we get
sin3A+sin3B+sin3C=3sinAsinBsinC.