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(A) Given determinant is $\Delta = \left|\begin{array}{lll}a & a^3 & a^4 \ b & b^3 & b^4 \ c & c^3 & c^4\end{array}ight|$.Taking $a, b, c$ common

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$abc(ab+bc+ca)$

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(A) Given determinant is $\Delta = \left|\begin{array}{lll}a & a^3 & a^4 \ b & b^3 & b^4 \ c & c^3 & c^4\end{array}ight|$.Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively,we get $\Delta = abc \left|\begin{array}{lll}1 & a^2 & a^3 \ 1 & b^2 & b^3 \ 1 & c^2 & c^3\end{array}ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = abc \left|\begin{array}{lll}1 & a^2 & a^3 \ 0 & b^2-a^2 & b^3-a^3 \ 0 & c^2-a^2 & c^3-a^3\end{array}ight|$.Taking $(b-a)$ common from $R_2$ and $(c-a)$ common from $R_3$:$\Delta = abc(b-a)(c-a) \left|\begin{array}{lll}1 & a^2 & a^3 \ 0 & b+a & b^2+ab+a^2 \ 0 & c+a & c^2+ac+a^2\end{array}ight|$.Expanding along $C_1$: $\Delta = abc(b-a)(c-a) [(c+a)(b^2+ab+a^2) - (b+a)(c^2+ac+a^2)]$.Simplifying the expression inside the bracket: $(cb^2+abc+a^2c+ab^2+a^2b+a^3) - (bc^2+abc+a^2b+ac^2+a^2c+a^3) = cb^2+ab^2-bc^2-ac^2 = b^2c-bc^2+ab^2-ac^2 = bc(b-c) + a(b-c)(b+c) = (b-c)(bc+ab+ac)$.Thus,$\Delta = abc(b-a)(c-a)(b-c)(ab+bc+ca) = abc(a-b)(b-c)(c-a)(ab+bc+ca)$.Comparing with $k(a-b)(b-c)(c-a)$,we get $k = abc(ab+bc+ca)$.

If $\left|\begin{array}{lll}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{array}\right|=k(a-b)(b-c)(c-a)$ then $k=$

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