How to calculate the determinants like these?

Question

I'm trying to solve this determinant question and I just can't understand how to approach this.

If $x^3$=1, then

$$\Delta=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$ equals

(A) $(cx^2+bx+a)\begin{vmatrix} 1 & b & c \\ x & c & a \\ x^2 & a & b \end{vmatrix}$ (B) $(cx^2+bx+a)\begin{vmatrix} x & b & c \\ 1 & c & a \\ x^2 & a & b \end{vmatrix}$

(C) $(cx^2+bx+a)\begin{vmatrix} x^2 & b & c \\ x & c & a \\ 1 & a & b \end{vmatrix}$ (D) $(cx^2+bx+a)\begin{vmatrix} 1 & b & c \\ x^2 & c & a \\ x & a & b \end{vmatrix}$

I don't understand the relevance of $x^3$. I can add all the rows and take out the common term, but I don't understand what I'm supposed to do here after that. Any hints would be really appreciated.

Answer 1

(D) is correct. Multiply $(cx^2+bx+a)$ in the first column, put 1 in place of $x^3$ (for example $cx^4=cx$...), then substract appropriate multiples of the second and third columns from the first column.

Answer 2

Since only first column varies in the given options, replace it with (p q r), multiply $(cx^2+bx+a)$ to the first column, we get

$$\begin{vmatrix} (cx^2+bx+a)p & b & c \\ (cx^2+bx+a)q & c & a \\ (cx^2+bx+a)r & a & b \end{vmatrix}$$

Now comparing with first column of determinant and using $x^3=1$, find the values of $p$ such that $ap=a$, $q$ such that $bxq=b$, $r$ such that $cx^2r=c$.

If you want to verify the answer, try and see if some column operations $C_1-C_ix^2-C_jx\to C_1$ for $i,j\in\{1,2\}$ takes you to the determinant given in question for the values of $p,q,r$ you just found.