If in a triangle ABC, \sin^{3}A+\sin^{3}B+\sin^{3}C=3 \sin A \sin B \sin C, then what is the value of the determinant \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} where a, b, c are sides of the triangle?

Explanation

The condition x^3+y^3+z^3=3xyz implies either x+y+z=0 or x=y=z. Since \sin A, \sin B, \sin C are positive in a triangle, their sum cannot be zero, hence \sin A = \sin B = \sin C. From the Sine Rule, this implies a=b=c (an equilateral triangle). The determinant of a matrix where all rows are cyclic permutations of equal elements evaluatesto 3abc - a^3 - b^3 - c^3. Substituting a=b=c gives 3a^3 - 3a^3 = 0.

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