Multiple choice

If $a, b, c$ are roots of the equation $Ax^{3}+Bx^{2}+C=0$ then the value of determinant $\begin{vmatrix} ab & bc & ca\ bc & ca & ab\ ca & ab & bc \end{vmatrix}$ equals to=?

  1. $A$
  2. $B$
  3. $C$
  4. $0$
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D Correct answer
Explanation

The determinant is a circulant determinant. For roots a, b, c of Ax^3 + Bx^2 + C = 0, the product abc = -C/A. The determinant evaluates to -(a^3 + b^3 + c^3 - 3abc)(abc). Since a, b, c are roots, the expression is related to the symmetric polynomials. Specifically, for this structure, the determinant is 0.

AI explanation

Expand the determinant using the standard rule. The expansion yields (ab)^3 + (bc)^3 + (ca)^3 - 3(ab)(bc)(ca). Factor this using the identity p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp). Here, the sum of the terms p + q + r evaluates to ab + bc + ca. By Vieta's formulas for the cubic Ax^3 + Bx^2 + C = 0, the sum of the pairwise products of the roots (ab + bc + ca) is zero. Because the first factor of the determinant's factored form is zero, the entire value of the determinant is 0.

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