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If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}}
{{q_1}}&{{q_2}}&{{q_3}} \\
{{q_2}}&{{q_3}}&{{q_1}} \\
{{q_3}}&{{q_1}}&{{q_2}}
\end{array}} \right|$ is
$1$
$4$
$16$
$0$
Solution
$\left|\begin{array}{lll}{\mathrm{q}_{1}} & {\mathrm{q}_{2}} & {\mathrm{q}_{3}} \\ {\mathrm{q}_{2}} & {\mathrm{q}_{3}} & {\mathrm{q}_{1}} \\ {\mathrm{q}_{3}} & {\mathrm{q}_{1}} & {\mathrm{q}_{2}}\end{array}\right| \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$
$=\left(\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}\right)\left|\begin{array}{lll}{1} & {\mathrm{q}_{2}} & {\mathrm{q}_{3}} \\ {1} & {\mathrm{q}_{3}} & {\mathrm{q}_{1}} \\ {1} & {\mathrm{q}_{1}} & {\mathrm{q}_{2}}\end{array}\right|$
$=0$ ($\because$ sum of roots is zero)