$\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
If $\omega $ is a complex cube root of unity, then the determinant $\left| {\,\begin{array}{*{20}{c}}2&{2\omega }&{ - {\omega ^2}}\\1&1&1\\1&{ - 1}&0\end{array}\,} \right| = $
Without expanding the determinant, prove that
$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$
Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$
If $\omega $is a cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right| = $