In a triangle $A B C$ ,if $s i n^{3} A + s i n^{3} B + s i n^{3} C = 3 s i n A s i n B s i n C$
then $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = ?$

0

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(a + b + c)^{3}

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(a + b + c) (a b + b c + c a)

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a b + b c + c a

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Solution

The correct option is A $0$
According to sine Rule,
$a = 2 R s i n A b = 2 R s i n B c = 2 R s i n C$

Now, $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = a b c - a^{3} - b^{3} + a b c + a b c - c^{3} \Rightarrow 3 a b c - (a^{3} + b^{3} + c^{3}) = 8 R^{3} [3 s i n A s i n B s i n C - (s i n^{3} A + s i n^{3} B + s i n^{3} C)]$

and it is given that $3 s i n A s i n B s i n C = s i n^{3} A + s i n^{3} B + s i n^{3} C$

$\Rightarrow 8 R^{3} [3 s i n A s i n B s i n C - (s i n^{3} A + s i n^{3} B + s i n^{3} C)] = 8 R^{3} \times 0 = 0$

Therefore, Answer is $A$

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In a triangle ABC,if sin3A+sin3B+sin3C=3sinAsinBsinC then ∣∣ ∣∣abcbcacab∣∣ ∣∣=?

In a triangle $A B C$ ,if $s i n^{3} A + s i n^{3} B + s i n^{3} C = 3 s i n A s i n B s i n C$
then $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = ?$