TITLE: Can this circulant determinant be zero?
QUESTION [7 upvotes]: The question is:
If $a,b,c$ are negative distinct real numbers,then the determinant $$ \begin{vmatrix}
a & b & c \\ 
b & c & a\\ 
c & a & b
\end{vmatrix} $$ is $$(a) \le 0 \quad (b) \gt 0 \quad (c) \lt 0 \quad (d) \ge 0 $$
My strategy: I identifed that the matrix is a circulant hence the determinant can be expressed in the form of $-(a^3 + b^3 + c^3 - 3abc)$ which implies that $-(a+b+c)\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$ hence $(-)(-)(+) \gt 0$ but the answers says it is $\ge 0$, so can we have three $a,b,c$ such that the answer is $0$ ?

REPLY [4 votes]: If you require $a, b, c$ distinct, no.  The term $a+b+c$ can't be $0$ as they are all the same sign and the other term is a sum of squares which can only be $0$ if $a=b=c$.

REPLY [4 votes]: Consider the equation:
$$(a^3 + b^3 + c^3)/3 = abc$$
as expressing the equality of the arithmetic and geometric means of $a^3,b^3,c^3$.  By a well known result this is only possible if the three cubes are equal.  So, no, we cannot get them distinct and have the determinant be zero.