If `|(x,y,0),(0,x,y),(y,0,x)| = 0`, then which one of the following is correct ?

Options

A. A. `(x)/(y)` is one of the cube roots of unity
B. B. x is one of the cube roots of unity
C. C. y is one of the cube roots of unity
D. D. `(x)/(y)` is one of the cube roots of -1.

Answer

To solve the given determinant problem, we need to find the determinant of the matrix:

\[ \begin{vmatrix} x & y & 0 \\ 0 & x & y \\ y & 0 & x \end{vmatrix} \]

and set it equal to zero.

### Step 1: Calculate the Determinant

We can use the method of cofactor expansion along the first row. The determinant can be calculated as follows:

\[ D = x \begin{vmatrix} x & y \\ 0 & x \end{vmatrix} - y \begin{vmatrix} 0 & y \\ y & x \end{vmatrix} + 0 \begin{vmatrix} 0 & x \\ y & 0 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

Now we calculate the two 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} x & y \\ 0 & x \end{vmatrix} = x \cdot x - y \cdot 0 = x^2 \]

2. For the second determinant:
\[ \begin{vmatrix} 0 & y \\ y & x \end{vmatrix} = 0 \cdot x - y \cdot y = -y^2 \]

### Step 3: Substitute Back into the Determinant

Substituting these back into the determinant calculation:

\[ D = x \cdot x^2 - y \cdot (-y^2) = x^3 + y^3 \]

### Step 4: Set the Determinant to Zero

According to the problem, we set the determinant equal to zero:

\[ x^3 + y^3 = 0 \]

### Step 5: Factor the Equation

We can factor this equation:

\[ x^3 = -y^3 \]

### Step 6: Divide by \(y^3\)

Dividing both sides by \(y^3\) (assuming \(y \neq 0\)) gives:

\[ \frac{x^3}{y^3} = -1 \]

### Step 7: Take the Cube Root

Taking the cube root of both sides, we have:

\[ \frac{x}{y} = \sqrt[3]{-1} \]

### Conclusion

The cube roots of \(-1\) are \(-1\), \(\frac{1}{2} + \frac{\sqrt{3}}{2} i\), and \(\frac{1}{2} - \frac{\sqrt{3}}{2} i\). Therefore, the correct option is:

**D: \( \frac{x}{y} \) is one of the cube roots of \(-1\)**.