If in a $Δ A B C, {sin}^{3} A + {sin}^{3} B + {sin}^{3} C = 3 sin A . sin B . sin C,$ then the value of the determinant
$∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣$ is

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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none of these

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Solution

The correct option is B $0$
${sin}^{3} A + {sin}^{3} B + {sin}^{3} C = 3 sin A . sin B . sin C$
Now from sine rule we get
$\frac{a}{s i n a} = \frac{b}{s i n b} = \frac{c}{s i n c}$

$a^{3} + b^{3} + c^{3} = 3 a b c$ ...(1)

$∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = a^{3} + b^{3} + c^{3} - 3 a b c$
Therefore, $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0$ ....[from (1)]

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If in a ΔABC,sin3A+sin3B+sin3C=3sinA.sinB.sinC, then the value of the determinant∣∣ ∣∣abcbcacab∣∣ ∣∣ is

The correct option is B 0sin3A+sin3B+sin3C=3sinA.sinB.sinCNow from sine rule we getasina=bsinb=csinca3+b3+c3=3abc ...(1) ∣∣ ∣∣abcbcacab∣∣ ∣∣=a3+b3+c3−3abcTherefore, ∣∣ ∣∣abcbcacab∣∣ ∣∣=0 ....[from (1)]

If in a $Δ A B C, {sin}^{3} A + {sin}^{3} B + {sin}^{3} C = 3 sin A . sin B . sin C,$ then the value of the determinant
$∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣$ is