Given x² + x + 1 = 0

[∵ (a + b)² = a² + 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

[∵ a² – b² = (a + b)(a – b)]

Thus, the roots of the given equation are.