left| begin{array}{ccc} 1 & 1 & 1 a & b & c | vedclass

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(C) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^3 & b^3 & c^3 \end{array} ight|$.If we put $a = b$,the first and second colu

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$(a + b + c)(a - b)(b - c)(c - a)$

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(C) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ a & b & c \ a^3 & b^3 & c^3 \end{array} ight|$.If we put $a = b$,the first and second columns become identical,so $\Delta = 0$. Thus,$(a - b)$ is a factor. Similarly,$(b - c)$ and $(c - a)$ are factors.The degree of the determinant is $1 + 1 + 3 = 5$ (or by inspection of the expansion,the highest power is $4$). Since the determinant is a homogeneous polynomial of degree $4$ in $a, b, c$,and we have already found factors $(a - b)(b - c)(c - a)$ of degree $3$,the remaining factor must be of degree $1$,which is $(a + b + c)$.Thus,$\Delta = k(a - b)(b - c)(c - a)(a + b + c)$.Comparing the coefficient of $bc^3$ on both sides,we find $k = 1$.Therefore,$\Delta = (a - b)(b - c)(c - a)(a + b + c)$.

$\left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array} \right| = $

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