Match each entry in List-I to the correct entry in List-II and choose the correct option. | List-I | List-II | | --- | --- | | (P) If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2} + x + 1 = 0$, then the quadratic equation with roots $\frac{1}{(\alpha+1)^{2026}}$ and $\frac{1}{(\beta+1)^{2026}}$ is | (1) $x^{2} + x + 1 = 0$ | | (Q) If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2} + x + 1 = 0$, then the quadratic equation with roots $\frac{1}{(\alpha+1)^{2027}}$ and $\frac{1}{(\beta+1)^{2027}}$ is | (2) $x^{2} - x + 1 = 0$ | | (R) If $\gamma$ and $\delta$ are the distinct roots of the equation $x^{2} - x + 1 = 0$, then the value of $\frac{1}{(\gamma-1)^{2026}} + \frac{1}{(\delta-1)^{2026}}$ is | (3) $x^{2} + x - 1 = 0$ | | (S) If p and r are the distinct roots of the equation $x^{2} + x - 1 = 0$, then the value of $\frac{1}{(p+1)^{3}} + \frac{1}{(r+1)^{3}}$ is | (4) $-1$ | | | (5) $-4$ |

$\text{(P),(Q): roots of } x^{2}+x+1=0$

$\text{(R): roots of } x^{2}-x+1=0$

$\text{(S): roots of } x^{2}+x-1=0$

$\text{To find: match (P),(Q),(R),(S) with (1)--(5)}$

$\alpha^{2}+\alpha+1=0 \Rightarrow \alpha+1 = -\alpha^{2}, \quad \alpha^{3}=1$

$(\alpha+1)^{2026} = (-\alpha^{2})^{2026} = \alpha^{4052} = \alpha^{2}$

$\frac{1}{(\alpha+1)^{2026}} = \frac{1}{\alpha^{2}} = \alpha$

$\text{Roots: } \alpha,\ \beta \Rightarrow x^{2}+x+1=0$

$(P)\to(1)$

$(\alpha+1)^{2027} = (-\alpha^{2})^{2027} = -\alpha^{4054} = -\alpha$

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

$\text{Sum } = -(\alpha^{2}+\beta^{2}) = -(-1) = 1$

$\text{Product } = (\alpha\beta)^{2} = 1$

$x^{2}-x+1=0 \Rightarrow (Q)\to(2)$

$\gamma^{2}-\gamma+1=0 \Rightarrow \gamma-1 = \gamma^{2}, \quad \gamma^{3}=-1,\ \gamma^{6}=1$

$(\gamma-1)^{2026} = \gamma^{4052} = \gamma^{2}$

$\frac{1}{(\gamma-1)^{2026}} = \gamma^{4}$

$\gamma^{4}+\delta^{4} = (\gamma^{2}+\delta^{2})^{2}-2(\gamma\delta)^{2}$

$= (-1)^{2}-2(1) = -1 \Rightarrow (R)\to(4)$

$p^{2}+p-1=0, \quad q = p+1$

$q^{2}-q-1=0$

$q^{2}=q+1,\quad \frac{1}{q}=q-1$

$\frac{1}{q^{3}} = (q-1)^{3} = 2q-3$

$\frac{1}{(p+1)^{3}}+\frac{1}{(r+1)^{3}} = 2(q_1+q_2)-6$

$q_1+q_2 = 1 \Rightarrow = -4 \Rightarrow (S)\to(5)$

$(P)\to(1),\ (Q)\to(2),\ (R)\to(4),\ (S)\to(5)$

$\text{Option } (3)$