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If a,b,ca,b,c are the roots of x3+px+q=0x^3+px+q=0, then the value of ∣abcbcacab∣\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} is equal to

Solution

✅ Correct Option: 1

This circulant determinant equals −(a3+b3+c3−3abc)=−(a+b+c)(a2+b2+c2−ab−bc−ca)-(a^3+b^3+c^3-3abc)=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca).

Since the cubic has no x2x^2 term, a+b+c=0a+b+c=0, so the whole product is zero.

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