If q_1, q_2, q_3 are roots of the equation x^ | vedclass

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(D) The given equation is $x^3 + 64 = 0$,which can be written as $x^3 = -64$.Let the roots of this equation be $q_1, q_2, q_3$. According to Vieta's f

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$0$

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(D) The given equation is $x^3 + 64 = 0$,which can be written as $x^3 = -64$.Let the roots of this equation be $q_1, q_2, q_3$. According to Vieta's formulas,the sum of the roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$ is given by $-b/a$.Here,the equation is $x^3 + 0x^2 + 0x + 64 = 0$,so the sum of the roots $q_1 + q_2 + q_3 = -0/1 = 0$.Now,consider the determinant $\Delta = \left| \begin{array}{ccc} q_1 & q_2 & q_3 \ q_2 & q_3 & q_1 \ q_3 & q_1 & q_2 \end{array} ight|$.Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\Delta = \left| \begin{array}{ccc} q_1 + q_2 + q_3 & q_2 & q_3 \ q_2 + q_3 + q_1 & q_3 & q_1 \ q_3 + q_1 + q_2 & q_1 & q_2 \end{array} ight|$Since $q_1 + q_2 + q_3 = 0$,the first column becomes zero:$\Delta = \left| \begin{array}{ccc} 0 & q_2 & q_3 \ 0 & q_3 & q_1 \ 0 & q_1 & q_2 \end{array} ight| = 0$.Thus,the value of the determinant is $0$.

If $q_1, q_2, q_3$ are roots of the equation $x^3 + 64 = 0$,then the value of $\left| \begin{array}{ccc} q_1 & q_2 & q_3 \\ q_2 & q_3 & q_1 \\ q_3 & q_1 & q_2 \end{array} \right|$ is

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