C1→C1+C2+C3=1+x+x2xx21+x+x21x1+x+x2x21=(1+x+x2)1xx211x1x21R2→R2−R1, R3→R3−R1=(1+x+x2)1xx201−xx(1−x)0x(x−1)1−x2=((1+x+x2)(1−x))21xx201x0−x1+x=(1+x+x2)2(1−x)2=((1+x+x2)(1−x))2=(1−x3)2

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$|\begin{matrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix}| =$

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${(1 - x^{3})}^{2}$

${(1 - x^{3})}^{3}$

$1 - x^{3}$

${(1 - x^{2})}^{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} C_{1} \to C_{1} + C_{2} + C_{3} \\ = |\begin{matrix} 1 + x + x^{2} & x & x^{2} \\ 1 + x + x^{2} & 1 & x \\ 1 + x + x^{2} & x^{2} & 1 \end{matrix}| \\ = (1 + x + x^{2}) |\begin{matrix} 1 & x & x^{2} \\ 1 & 1 & x \\ 1 & x^{2} & 1 \end{matrix}| \\ R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} \\ = (1 + x + x^{2}) |\begin{matrix} 1 & x & x^{2} \\ 0 & 1 - x & x (1 - x) \\ 0 & x (x - 1) & 1 - x^{2} \end{matrix}| \\ = {((1 + x + x^{2}) (1 - x))}^{2} |\begin{matrix} 1 & x & x^{2} \\ 0 & 1 & x \\ 0 & - x & 1 + x \end{matrix}| = {(1 + x + x^{2})}^{2} {(1 - x)}^{2} \\ = {((1 + x + x^{2}) (1 - x))}^{2} \\ = {(1 - x^{3})}^{2} \end{array}$