Roots of the equation `x^(2017) + x^(2018) +1=0` are

Options

A. `-1, omega`
B. `1, omega`
C. ` -1 , omega^(2)`
D. ` omega, omega^(2)`

Answer

To find the roots of the equation \( x^{2017} + x^{2018} + 1 = 0 \), we can start by rearranging the equation:

1. **Rearranging the Equation**:
\[ x^{2018} + x^{2017} + 1 = 0 \]
This can be rewritten as:
\[ x^{2018} + x^{2017} = -1 \]

2. **Factoring the Left Side**:
We can factor out \( x^{2017} \):
\[ x^{2017}(x + 1) + 1 = 0 \]
This implies:
\[ x^{2017}(x + 1) = -1 \]

3. **Substituting \( x = \omega \)**:
We know that \( \omega \) is a primitive cube root of unity, where \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). We can test if \( x = \omega \) is a root:
\[ \omega^{2017} + \omega^{2018} + 1 = 0 \]
Since \( \omega^3 = 1 \):
\[ \omega^{2017} = \omega^{2017 \mod 3} = \omega^1 = \omega \]
\[ \omega^{2018} = \omega^{2018 \mod 3} = \omega^2 \]
Thus:
\[ \omega + \omega^2 + 1 = 0 \]
This confirms that \( x = \omega \) is indeed a root.

4. **Substituting \( x = \omega^2 \)**:
Similarly, we can check \( x = \omega^2 \):
\[ (\omega^2)^{2017} + (\omega^2)^{2018} + 1 = 0 \]
Again using \( \omega^3 = 1 \):
\[ (\omega^2)^{2017} = \omega^{2 \cdot 2017 \mod 3} = \omega^2 \]
\[ (\omega^2)^{2018} = \omega^{2 \cdot 2018 \mod 3} = \omega \]
Thus:
\[ \omega^2 + \omega + 1 = 0 \]
This confirms that \( x = \omega^2 \) is also a root.

5. **Conclusion**:
The roots of the equation \( x^{2017} + x^{2018} + 1 = 0 \) are \( \omega \) and \( \omega^2 \).

**Final Answer**: The roots of the equation are \( \omega \) and \( \omega^2 \).

---