If `q_(1),q_(2),q_(3)` ar rootsof the equation `x^(3)+64=0` then value of `|{:(q_(1),q_(2),q_(3)),(q_(2),q_(3),q_(1)),(q_(3),q_(1),q_(2)):}|`

Options

A. 1
B. 4
C. 16
D. 0

Explanation

To solve the problem, we need to evaluate the determinant of the following matrix formed by the roots \( q_1, q_2, q_3 \) of the equation \( x^3 + 64 = 0 \):

\[ D = \begin{vmatrix} q_1 & q_2 & q_3 \\ q_2 & q_3 & q_1 \\ q_3 & q_1 & q_2 \end{vmatrix} \]

### Step 1: Find the roots of the equation
The given equation is:

\[ x^3 + 64 = 0 \]

This can be rewritten as:

\[ x^3 = -64 \]

Taking the cube root, we find:

\[ x = -4 \]

The roots of the equation are the cube roots of \(-64\). The three roots are:

\[ q_1 = -4, \quad q_2 = 4\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right), \quad q_3 = 4\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) \]

Calculating these, we have:

\[ q_2 = 4\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -2 + 2i\sqrt{3} \]
\[ q_3 = 4\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -2 - 2i\sqrt{3} \]

### Step 2: Calculate the determinant
Now we need to calculate the determinant \( D \):

\[ D = \begin{vmatrix} q_1 & q_2 & q_3 \\ q_2 & q_3 & q_1 \\ q_3 & q_1 & q_2 \end{vmatrix} \]

### Step 3: Simplify the determinant
We can simplify the determinant by adding all rows together. Let's denote the sum of the roots:

\[ S = q_1 + q_2 + q_3 \]

Since \( q_1, q_2, q_3 \) are the roots of the polynomial \( x^3 + 64 = 0 \), we can find \( S \):

Using Vieta's formulas, the sum of the roots \( S \) is given by:

\[ S = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{0}{1} = 0 \]

### Step 4: Substitute back into the determinant
Now, we substitute \( S \) back into the determinant:

\[ D = \begin{vmatrix} S & q_2 & q_3 \\ q_2 & q_3 & S \\ q_3 & S & q_2 \end{vmatrix} = \begin{vmatrix} 0 & q_2 & q_3 \\ q_2 & q_3 & 0 \\ q_3 & 0 & q_2 \end{vmatrix} \]

Since the first row contains a zero, the determinant simplifies to zero:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]