(a) $\left[ {\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} – 1}\\b&{{b^3}}&{{b^4} – 1}\\c&{{c^3}}&{{c^4} – 1}\end{array}} \right]\, = 0$

or $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4}}\\b&{{b^3}}&{{b^4}}\\c&{{c^3}}&{{c^4}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ – 1}\\b&{{b^3}}&{ – 1}\\c&{{c^3}}&{ – 1}\end{array}\,} \right| = 0$

or $abc{\rm{ }}\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ – 1}\\{a – b}&{{a^3} – {b^3}}&0\\{a – c}&{{a^3} – {c^3}}&0\end{array}\,} \right|\, = 0$

or $abc{\rm{ }}\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\0&{{a^2} – {b^2}}&{{a^3} – {b^3}}\\0&{{a^2} – {c^2}}&{{a^3} – {c^3}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ – 1}\\{a – b}&{{a^3} – {b^3}}&0\\{(a – c)}&{({a^3} – {c^3})}&0\end{array}\,} \right|\, = 0$

or $(abc)\,(a – b)\,(a – c)\,\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\0&{a + b}&{{a^2} + {b^2} + ab}\\0&{a + c}&{{a^2} + {c^2} + ac}\end{array}\,} \right|\, + $

$(a – b)\,(a – c)\,\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ – 1}\\1&{{a^2} + {b^2} + ab}&0\\1&{{a^2} + {c^2} + ac}&0\end{array}\,} \right|$

or $(a – b)\,(a – c)\,[(abc)[(a + b)\,({a^2} + {c^2} + ac) – $

$(a + c)({a^2} + {b^2} + ab)]] + ( – 1)\,(a – b)\,(a – c)$

$[{a^2} + {c^2} + ac – {a^2} – {b^2} – ab] = 0$

= $(abc)\,[(a – b)\,(a – c)\,(c – b)(ac + ab + bc)]$

$ + ( – 1)(a – b)(a – c)(c – b)(a + b + c) = 0$

$ \Rightarrow $ $(abc)\,(ac + ab + bc) = a + b + c$.