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 $ $(abc)\,(ac + ab + bc) = a + b + c$.
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$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :