If w is a complex cube root of unity.
$\begin{vmatrix}a &b& c\\ b&c& a\\c&a&b\end{vmatrix}=-(a+b+c) (a + bk + ck^2) (a + bk^2 + ck)$, then k equals
Answer & explanation
Correct answer: option 3
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 $C_1→C_1 + C_2+C_3$]
$=(a+b+c)\begin{vmatrix}1 &b& c\\ 1&c& a\\1&a&b\end{vmatrix}$
$=(a+b+c)\begin{vmatrix}1 &b& c\\ 0&c-b& a-c\\0&a-b&b-c\end{vmatrix}$ [Applying $R_2 → R_2 −R_1, R_3→R_3-R_1$
$=(a+b+c)\begin{vmatrix}c-b& a-c\\a-b&b-c\end{vmatrix}$
$=(a+b+c) (-b^2 - c^2-a^2 + ac + ab + bc - 2bc)$
$=-(a+b+c) (a^2 + b^2 + c^2-ab-bc-ca)$
$=−(a+b+c) (a + bw + cw^2)(a + bw^2 + cw)$
$∴ k = w$