If `|{:(a,b,0),(0,a,b),(b,0,a):}|=0`, then which one of the following is correct ?

Options

A. a/b is one of the cube roots of unity
B. a is one of the cube roots of unity
C. b is one of the cube roots of unity
D. a/b is one of the cube roots of `-1`

Explanation

To solve the determinant equation given by:

\[ \begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Calculate the Determinant

We will calculate the determinant using the method of cofactor expansion along the first row.

\[ D = a \begin{vmatrix} a & b \\ 0 & a \end{vmatrix} - b \begin{vmatrix} 0 & b \\ b & a \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & a \\ b & 0 \end{vmatrix} \]

### Step 2: Evaluate the 2x2 Determinants

Now we calculate the two 2x2 determinants:

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

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

### Step 3: Substitute Back into the Determinant Expression

Substituting these values back into the determinant expression, we have:

\[ D = a(a^2) - b(-b^2) = a^3 + b^3 \]

### Step 4: Set the Determinant Equal to Zero

Since we are given that the determinant equals zero, we set up the equation:

\[ a^3 + b^3 = 0 \]

### Step 5: Factor the Expression

We can factor \(a^3 + b^3\) using the identity \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\):

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 0 \]

### Step 6: Analyze the Factors

This gives us two cases:

1. \(a + b = 0\)
2. \(a^2 - ab + b^2 = 0\)

### Step 7: Solve the Cases

1. From \(a + b = 0\), we get \(a = -b\).
2. The second equation \(a^2 - ab + b^2 = 0\) can be analyzed further.

### Step 8: Use the Second Equation

The equation \(a^2 - ab + b^2 = 0\) can be rearranged to find the relationship between \(a\) and \(b\):

\[ a^2 + b^2 = ab \]

### Step 9: Divide by \(b^2\) (assuming \(b \neq 0\))

Dividing the entire equation by \(b^2\):

\[ \left(\frac{a}{b}\right)^2 + 1 = \frac{a}{b} \]

Let \(x = \frac{a}{b}\):

\[ x^2 - x + 1 = 0 \]

### Step 10: Find the Roots

The roots of this quadratic equation can be found using the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]

### Conclusion

The roots \(x = \frac{a}{b}\) are complex numbers, specifically the cube roots of unity. Therefore, we conclude that:

\[ \frac{a}{b} \text{ is one of the cube roots of } -1. \]

### Final Answer

Thus, the correct option is that \( \frac{a}{b} \) is one of the cube roots of \(-1\).

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