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(D) Given determinant: $\Delta = \left|\begin{array}{ccc}1 & 1 & 1 \ a^2 & b^2 & c^2 \ a^3 & b^3 & c^3\end{array}ight|$Applying column operations

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$(a-b)(b-c)(c-a)(ab+bc+ca)$

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(D) Given determinant: $\Delta = \left|\begin{array}{ccc}1 & 1 & 1 \ a^2 & b^2 & c^2 \ a^3 & b^3 & c^3\end{array}ight|$Applying column operations $C_1 ightarrow C_1 - C_2$ and $C_3 ightarrow C_3 - C_2$:$\Delta = \left|\begin{array}{ccc}0 & 1 & 0 \ a^2-b^2 & b^2 & c^2-b^2 \ a^3-b^3 & b^3 & c^3-b^3\end{array}ight|$Taking common factors $(a-b)$ from $C_1$ and $(c-b)$ from $C_3$:$\Delta = (a-b)(c-b) \left|\begin{array}{ccc}0 & 1 & 0 \ a+b & b^2 & c+b \ a^2+ab+b^2 & b^3 & c^2+bc+b^2\end{array}ight|$Applying $C_3 ightarrow C_3 - C_1$:$\Delta = (a-b)(c-b) \left|\begin{array}{ccc}0 & 1 & 0 \ a+b & b^2 & c-a \ a^2+ab+b^2 & b^3 & c^2-a^2+bc-ab\end{array}ight|$Since $c^2-a^2+bc-ab = (c-a)(c+a) + b(c-a) = (c-a)(a+b+c)$,we take $(c-a)$ common from $C_3$:$\Delta = (a-b)(c-b)(c-a) \left|\begin{array}{ccc}0 & 1 & 0 \ a+b & b^2 & 1 \ a^2+ab+b^2 & b^3 & a+b+c\end{array}ight|$Expanding along $R_1$:$\Delta = (a-b)(c-b)(c-a) \cdot (-1) \cdot [(a+b)(a+b+c) - (a^2+ab+b^2)]$$\Delta = (a-b)(b-c)(c-a) [a^2+ab+ac+ab+b^2+bc - a^2-ab-b^2]$$\Delta = (a-b)(b-c)(c-a)(ab+bc+ca)$.

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

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