Match each entry in List-I to the correct entry in List-II and choose the correct option.

Given $$x^2 + x + 1 = 0 \implies \alpha = \omega, \ \beta = \omega^2$$

$$\text{Since } 1 + \omega + \omega^2 = 0 \implies \alpha + 1 = -\omega^2, \ \beta + 1 = -\omega$$

Evaluating entry (P):

$$\text{Roots: } \frac{1}{(\alpha+1)^{2026}} = \frac{1}{(-\omega^2)^{2026}} = \frac{1}{\omega^{4052}} = \frac{1}{\omega^2} = \omega$$

$$\frac{1}{(\beta+1)^{2026}} = \frac{1}{(-\omega)^{2026}} = \frac{1}{\omega^{2026}} = \frac{1}{\omega} = \omega^2$$

$$\text{Equation with roots } \omega, \omega^2 \text{ is } x^2 + x + 1 = 0 \implies \text{(P)} \rightarrow \text{(1)}$$

Evaluating entry (Q):

$$\text{Roots: } \frac{1}{(\alpha+1)^{2027}} = \frac{1}{(-\omega^2)^{2027}} = \frac{1}{-\omega^{4054}} = \frac{1}{-\omega} = -\omega^2$$

$$\frac{1}{(\beta+1)^{2027}} = \frac{1}{(-\omega)^{2027}} = \frac{1}{-\omega^{2027}} = \frac{1}{-\omega^2} = -\omega$$

$$\text{Equation with roots } -\omega, -\omega^2 \text{ is } x^2 - (-\omega - \omega^2)x + (-\omega)(-\omega^2) = 0 \implies x^2 - x + 1 = 0 \implies \text{(Q)} \rightarrow \text{(2)}$$

Evaluating entry (R):

$$\text{For } x^2 - x + 1 = 0 \implies \gamma = -\omega, \ \delta = -\omega^2$$

$$\gamma - 1 = -\omega - 1 = \omega^2, \quad \delta - 1 = -\omega^2 - 1 = \omega$$

$$\frac{1}{(\gamma-1)^{2026}} + \frac{1}{(\delta-1)^{2026}} = \frac{1}{(\omega^2)^{2026}} + \frac{1}{(\omega)^{2026}} = \frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1 \implies \text{(R)} \rightarrow \text{(4)}$$

Evaluating entry (S):

$$\text{For } x^2 + x - 1 = 0 \implies p^2 + p = 1 \implies p + 1 = \frac{1}{p}, \quad r + 1 = \frac{1}{r}$$

$$\frac{1}{(p+1)^3} + \frac{1}{(r+1)^3} = p^3 + r^3 = (p+r)(p^2 - pr + r^2) = (p+r)((p+r)^2 - 3pr)$$

$$\text{Since } p+r = -1, \ pr = -1:$$ 

$$= (-1)((1) - 3(-1)) = (-1)(4) = -4 \implies \text{(S)} \rightarrow \text{(5)}$$