If `omega ` is a cube root of unity and ` omega ^(n) + omega ^(2n) = - 1 ` then the integer n is of the form km + l where (k , l) = . . .

Answer

To solve the problem, we need to analyze the equation given:

\[ \omega^n + \omega^{2n} = -1 \]

where \(\omega\) is a cube root of unity. The cube roots of unity are:

\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \]
\[ \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \]
\[ \omega^3 = 1 \]

### Step 1: Use properties of cube roots of unity
From the properties of cube roots of unity, we know:

\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]

This implies:

\[ \omega + \omega^2 = -1 \]

### Step 2: Substitute \(n\) in terms of \(3m\), \(3m+1\), and \(3m+2\)
We can express \(n\) in the form \(n = 3m + k\) where \(k = 0, 1, 2\).

#### Case 1: \(n = 3m\)
\[ \omega^{3m} + \omega^{2(3m)} = 1 + 1 = 2 \quad \text{(not equal to -1)} \]

#### Case 2: \(n = 3m + 1\)
\[ \omega^{3m+1} + \omega^{2(3m+1)} = \omega + \omega^2 = -1 \quad \text{(this works)} \]

#### Case 3: \(n = 3m + 2\)
\[ \omega^{3m+2} + \omega^{2(3m+2)} = \omega^2 + \omega = -1 \quad \text{(this also works)} \]

### Conclusion
The integer \(n\) can be expressed in the form:

\[ n = 3m + 1 \quad \text{or} \quad n = 3m + 2 \]

Thus, we can conclude that \(n\) is of the form \(km + l\) where \(k = 3\) and \(l = 1\) or \(2\).

### Final Answer
The integer \(n\) is of the form \(3m + l\) where \(l = 1 \text{ or } 2\).