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$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $
${a^2} + {b^2} + {c^2}$
$(a + b)\,(b + c)\,(c + a)$
$(a - b)(b - c)(c - a)$
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Solution
(c) $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right|\, = \left| {\,\begin{array}{*{20}{c}}0&{a – b}&{{a^2} – {b^2}}\\0&{b – c}&{{b^2} – {c^2}}\\1&c&{{c^2}}\end{array}\,} \right|,$ by $\begin{array}{l}{R_1} \to {R_1} – {R_2}\\{R_2} \to {R_2} – {R_3}\end{array}$
= $(a – b)\,(b – c)\,\left| {\,\begin{array}{*{20}{c}}0&1&{a + b}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}\,} \right|$
= $(a – b)\,\,(b – c)\,\left| {\,\begin{array}{*{20}{c}}0&0&{a – c}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}\,} \right|$, by ${R_1} \to {R_1} – {R_2}$
= $(a – b)\,(b – c)\,(a – c)\,\left| {\,\begin{array}{*{20}{c}}0&0&1\\0&1&{b + c}\\1&c&{{c^2}}\end{array}\,} \right|$
= $(a – b)\,(b – c)\,(a – c)\,.\,( – 1) = (a – b)\,(b – c)\,(c – a)$.