If α, β are roots of x² - x + 1 = 0, find the value of α²⁰²⁶ + β²⁰²⁶ — JEE Mathematics

The roots of $x^2 - x + 1 = 0$ are given by:

$$x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{3}i}{2}$$

These are the negative complex cube roots of unity. Let $\alpha = -\omega$ and $\beta = -\omega^2$, where $\omega = \frac{-1 + \sqrt{3}i}{2}$.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

Now evaluate the expression:

$$\alpha^{2026} + \beta^{2026} = (-\omega)^{2026} + (-\omega^2)^{2026}$$

Since $2026$ is an even exponent, the negative signs become positive:

$$\omega^{2026} + \omega^{4052}$$

Simplify the powers of $\omega$ using $\omega^3 = 1$:

$$2026 = 3 \times 675 + 1 \implies \omega^{2026} = \omega^1 = \omega$$

$$4052 = 3 \times 1350 + 2 \implies \omega^{4052} = \omega^2$$

Thus:

$$\alpha^{2026} + \beta^{2026} = \omega + \omega^2 = -1$$

Answer: $-1$