If \( a,b,c \) are the roots of \( x^{3}+px+q=0 \), then the value of \( \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b\end{vmatrix} \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
The cubic has no \( x^{2} \) term, so the sum of the roots is
\( \displaystyle a+b+c=-\frac{\text{coefficient of }x^{2}}{\text{coefficient of }x^{3}} \)
\( \displaystyle =0 \)
In the determinant apply \( C_{1}\rightarrow C_{1}+C_{2}+C_{3} \):
\( \displaystyle \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b\end{vmatrix}=\begin{vmatrix} a+b+c & b & c\\ a+b+c & c & a\\ a+b+c & a & b\end{vmatrix} \)
Every entry of the first column is a + b + c = 0, so the determinant is 0.