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$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $
$4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
$3\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
$2\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
None of these
Solution
(a) Apply ${R_2} – {R_3}$ and note that ${(x + y)^2} – {(x – y)^2} = 4xy$
$\therefore $ $\Delta = 4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\{{{(a – 1)}^2}}&{{{(b – 1)}^2}}&{{{(c – 1)}^2}}\end{array}\,} \right|$
= $4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
{Applying ${R_3} – ({R_1} – 2{R_2}) $}
