(x² + x + 1)k = (x² – x + 1)

(k – 1)x² + (k + 1)x + (k – 1) = 0

For roots of quadratic equation real

Case I : a ≠ 0 and D ≥ 0

k – 1 ≠ 0 ⇒ k ≠ 1

-3k² + 10k – 3 ≥ 0

3k² – 10k + 3 ≤ 0

or

or

Case II : a = 0

k – 1 = 0 ⇒ k = 1

At k = 1, 2x = 0 ⇒ x = 0 is real

So, k = 1 is also count in answer.

Then, final answer is k ∈ [1/3, 3]