If $\omega $ is a complex cube root of unity and
$\begin{vmatrix}a & b & c\\ b & c & a\\c & a & b \end{vmatrix} = - ( a+b +c) ( a+ bk + ck^2)(a+bk^2+ck),$
then $1+k+k^2$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : 0
We have,
$\begin{vmatrix}a & b & c\\ b & c & a\\c & a & b \end{vmatrix}$
$= \begin{vmatrix}a+b+c & b & c\\a+ b+c & c & a\\a+ b+c & a & b \end{vmatrix}$ [Applying $C_1→C_1+C_2+C_3$]
$=(a+b+c) \begin{vmatrix}1 & b & c\\ 1 & c & a\\1 & a & b \end{vmatrix}$
$=(a+b+c) \begin{vmatrix}1 & b & c\\ 0 & c-b & a-c\\0 & a-b & b-c \end{vmatrix}$ $\begin{bmatrix}Applying \, R_2→R_2-R_1\\R_3→R_3-R_1\end{bmatrix}$
$=(a+b+c) \begin{vmatrix}c-b & a-c\\ a-b & b-c \end{vmatrix}$
$= ( a+b+c) (-b^2 -c^2 -a^2 +ac+ab + bc -2bc)$
$=-(a+b+c)(a^2 +b^2 +c^2 -ab -bc - ca)$
$= - (a + b +c ) ( a+ b \omega + c \omega^2 ) (a + b \omega^2 + c \omega )$
$∴k = \omega $
Hence, $1+k + k^2 = 1+ \omega + \omega^2 = 0.$