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# If $a,b,c$ are all different and $\begin{vmatrix}a&a^3&a^4-1\\b&b^3&b^4-1\\c&c^3&c^4-1\end{vmatrix}=0$ then the value of $abc(ab+bc+ca)$ is

$\begin{array}{1 1}(A)\;a+b+c&(B)\;0\\(C)\;a^2+b^2+c^2&(D)\;a^2-b^2+c^2\end{array}$

$\begin{vmatrix}a&a^3&a^4-1\\b&b^3&b^4-1\\c&c^3&c^4-1\end{vmatrix}=0$
$\Rightarrow \begin{vmatrix}a&a^3&a^4\\b&b^3&b^4\\c&c^3&c^4\end{vmatrix}-\begin{vmatrix}a&a^3&1\\b&b^3&1\\c&c^3&1\end{vmatrix}=0$
$\Rightarrow abc\begin{vmatrix}1&a^2&a^3\\1&b^2&b^3\\1&c^2&c^3\end{vmatrix}-\begin{vmatrix}1&a&a^2\\1&b&b^3\\1&c&c^3\end{vmatrix}=0$
$\Rightarrow abc(a-b)(b-c)(c-a)(ab+bc+ca)-(a-b)(b-c)(c-a)(a+b+c)=0$
$\Rightarrow (a-b)(b-c)(c-a)\times [abc(ab+bc+ca)-(a+b+c)]=0$
Hence (B) is the correct answer.