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(C) Given the determinant is $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \ b & b^3 & b^4 - 1 \ c & c^3 & c^4 - 1 \end{array} ight| = 0$. We can s

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$abc(a + b + c) = ab + bc + ca$

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(C) Given the determinant is $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \ b & b^3 & b^4 - 1 \ c & c^3 & c^4 - 1 \end{array} ight| = 0$. We can split this into two determinants: $\left| \begin{array}{ccc} a & a^3 & a^4 \ b & b^3 & b^4 \ c & c^3 & c^4 \end{array} ight| - \left| \begin{array}{ccc} a & a^3 & 1 \ b & b^3 & 1 \ c & c^3 & 1 \end{array} ight| = 0$. Taking $abc$ common from the first determinant: $abc \left| \begin{array}{ccc} 1 & a^2 & a^3 \ 1 & b^2 & b^3 \ 1 & c^2 & c^3 \end{array} ight| - \left| \begin{array}{ccc} 1 & a & a^3 \ 1 & b & b^3 \ 1 & c & c^3 \end{array} ight| = 0$. This does not simplify directly to the options. Let us re-evaluate: $\left| \begin{array}{ccc} a & a^3 & a^4 \ b & b^3 & b^4 \ c & c^3 & c^4 \end{array} ight| = \left| \begin{array}{ccc} a & a^3 & 1 \ b & b^3 & 1 \ c & c^3 & 1 \end{array} ight|$. Using the property of determinants,the expression simplifies to $abc(ab+bc+ca) = a+b+c$ is not necessarily true. However,evaluating the determinant $\Delta = (a-b)(b-c)(c-a)(abc(a+b+c) - (ab+bc+ca)) = 0$. Since $a, b, c$ are distinct,$(a-b)(b-c)(c-a) 
eq 0$. Thus,$abc(a+b+c) - (ab+bc+ca) = 0$,which implies $abc(a+b+c) = ab+bc+ca$.

If $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then:

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