left| {,begin{array}{*{20}{c}}1&a&{{a^2} - bc | vedclass

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(A) Let $\Delta = \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \ 1 & b & {{b}^{2}}-ac \ 1 & c & {{c}^{2}}-ab \end{matrix} ight|$.Applying row operat

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(A) Let $\Delta = \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \ 1 & b & {{b}^{2}}-ac \ 1 & c & {{c}^{2}}-ab \end{matrix} ight|$.Applying row operations $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{matrix} 0 & a-b & (a^2-bc) - (b^2-ac) \ 0 & b-c & (b^2-ac) - (c^2-ab) \ 1 & c & {{c}^{2}}-ab \end{matrix} ight|$Simplify the third column elements:$(a^2-bc) - (b^2-ac) = (a^2-b^2) + (ac-bc) = (a-b)(a+b) + c(a-b) = (a-b)(a+b+c)$$(b^2-ac) - (c^2-ab) = (b^2-c^2) + (ab-ac) = (b-c)(b+c) + a(b-c) = (b-c)(a+b+c)$Substituting these back:$\Delta = \left| \begin{matrix} 0 & a-b & (a-b)(a+b+c) \ 0 & b-c & (b-c)(a+b+c) \ 1 & c & {{c}^{2}}-ab \end{matrix} ight|$Taking $(a-b)$ common from $R_1$ and $(b-c)$ common from $R_2$:$\Delta = (a-b)(b-c) \left| \begin{matrix} 0 & 1 & a+b+c \ 0 & 1 & a+b+c \ 1 & c & {{c}^{2}}-ab \end{matrix} ight|$Since $R_1$ and $R_2$ are identical,the value of the determinant is $0$.

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

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