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Question

If for a triangle ABC,
∣ ∣abcbcacab∣ ∣=0
then sin3A+sin3B+sin3C=

A
sinA+sinB+sinC
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B
3sinAsinBsinC
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C
sin3A+sin3B+sin3C
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D
sin3Asin3Bsin3C
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Solution

The correct option is B 3sinAsinBsinC
Δ=∣ ∣abcbcacab∣ ∣=(a+b+c)∣ ∣1bc1ca1ab∣ ∣
[applying C1C1+C2+C3 and taking (a+b+c) common from C1]=(a+b+c)∣ ∣1bc0cbac0abbc∣ ∣[applying R2R2R1 and R3R3R1]=(a+b+c)[(cb)(bc)(ab)(ac)]=(a+b+c)[ab+bc+caa2b2c2]=(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.

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