If $a,b,c$ are unequal what is the condition that the value of the following determinant is zero $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|$

  • [IIT 1985]
  • A

    $1 + abc = 0$

  • B

    $a + b + c + 1 = 0$

  • C

    $(a - b)(b - c)(c - a) = 0$

  • D

    None of these

Similar Questions

If $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :

If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

The value of $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$is

$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$