Solve the equation $\begin{vmatrix} a-x & c & b \ c & b-x & a \ b & a & c-x \end{vmatrix}=0$ where $a+b+c\neq 0$.
- $x=a+b+c,\pm \sqrt { \cfrac { 1 }{ 2 } \left\{ { \left( a-b \right) }^{ 2 }-{ \left( b-c \right) }^{ 2 }-{ \left( c-a \right) }^{ 2 } \right\} } $
- $x=a+b+c,\pm \sqrt { \cfrac { 1 }{ 2 } \left\{ { \left( a-b \right) }^{ 2 }+{ \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 } \right\} } $
- $0$
- $1$
The determinant of this circulant matrix is -(a+b+c)(x^2 - (a+b+c)x + (ab+bc+ca - a^2-b^2-c^2)). Setting to 0 gives x = a+b+c or x^2 - (a+b+c)x + (ab+bc+ca - a^2-b^2-c^2) = 0. The roots of the quadratic part are (a+b+c)/2 +/- sqrt((a+b+c)^2/4 - (ab+bc+ca - a^2-b^2-c^2)). This simplifies to the expression in option B.
Add the second and third rows to the first row to factor out (a + b + c - x) from the determinant. The determinant becomes (a + b + c - x) multiplied by a new matrix. By applying the matrix determinant lemma or row and column operations to simplify the remaining matrix, its determinant part reduces to 1/2[(a-b)^2 + (b-c)^2 + (c-a)^2] - x^2. Setting the overall determinant to zero gives the linear root x = a + b + c and the quadratic roots x = +/- sqrt(1/2[(a-b)^2 + (b-c)^2 + (c-a)^2]).