If `omega` is a cube root of unity, then a root of the following equation is
`|(x+1,omega,omega^2),(omega,x+omega^2,1),(omega^2,1,x+omega)|` = 0

Options

A. x = 1
B. `x=omega`
C. `x=omega^2`
D. x = 0

Answer

To solve the problem, we need to find the root of the determinant equation given by:

\[ \left| \begin{array}{ccc} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{array} \right| = 0 \]

Where \(\omega\) is a cube root of unity. The cube roots of unity satisfy the equation \(1 + \omega + \omega^2 = 0\).

### Step 1: Apply Row Operations
We can simplify the determinant by applying row operations. Specifically, we will add the second and third rows to the first row.

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives us:

\[ R_1 = (x + 1 + \omega + \omega^2, \omega + (x + \omega^2) + 1, \omega^2 + 1 + (x + \omega)) \]

Using the property of cube roots of unity, we know that \(\omega + \omega^2 = -1\). Thus:

- For the first element of \(R_1\):
\[ x + 1 + \omega + \omega^2 = x + 1 - 1 = x \]

- For the second element of \(R_1\):
\[ \omega + (x + \omega^2) + 1 = \omega + x - 1 + 1 = x + \omega \]

- For the third element of \(R_1\):
\[ \omega^2 + 1 + (x + \omega) = \omega^2 + 1 + x - 1 = x + \omega^2 \]

So the new determinant becomes:

\[ \left| \begin{array}{ccc} x & x + \omega & x + \omega^2 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = 0 \]

### Step 2: Factor Out \(x\)
Now, we can factor \(x\) out of the first row:

\[ = x \left| \begin{array}{ccc} 1 & 1 + \frac{\omega}{x} & 1 + \frac{\omega^2}{x} \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = 0 \]

This gives us two cases to consider:

1. \(x = 0\)
2. The determinant must equal zero.

### Step 3: Solve the Determinant
Now we need to solve the determinant:

\[ \left| \begin{array}{ccc} 1 & 1 + \frac{\omega}{x} & 1 + \frac{\omega^2}{x} \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = 0 \]

However, since we are only interested in finding a single root, we can conclude that one root is:

\[ x = 0 \]

### Conclusion
Thus, the root of the equation is:

\[ \boxed{0} \]