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(C) Let $A = \begin{bmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{bmatrix}$. We are given $|A| = 5$. The second determinant is $D = \left| \

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$25$

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(C) Let $A = \begin{bmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{bmatrix}$. We are given $|A| = 5$. The second determinant is $D = \left| \begin{matrix} bc(c-b) & ac(a-c) & ab(b-a) \ (b-c)(b+c) & (c-a)(c+a) & (a-b)(a+b) \ -(b-c) & -(c-a) & -(a-b) \end{matrix} ight|$. Factoring out $(b-c)$ from $R_1$,$(c-a)$ from $R_2$,and $(a-b)$ from $R_3$ is not direct,but we can observe the structure. The given determinant $D$ is the square of the original determinant $|A|$ because it represents the determinant of the adjugate or a related cofactor matrix transformation. Specifically,$D = |A|^2 = 5^2 = 25$.

If $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{matrix} \right| = 5$,then $\left| \begin{matrix} bc^2 - b^2c & a^2c - ac^2 & ab^2 - ba^2 \\ b^2 - c^2 & c^2 - a^2 & a^2 - b^2 \\ c - b & a - c & b - a \end{matrix} \right|$ is equal to

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