Discussion

Answer: C

The correct match is (P) → (1), (Q) → (2), (R) → (4), (S) → (5).

The roots of \(x^{2} + x + 1 = 0\) are the cube roots of unity \(\omega, \omega^{2}\), with \(\omega^{3} = 1\) and \(1 + \omega + \omega^{2} = 0\). So \(\omega + 1 = -\omega^{2}\) and \(\omega^{2} + 1 = -\omega\).

(P) \((\omega + 1)^{2026} = \omega^{4052} = \omega^{2}\) (since \(4052 = 3 \times 1350 + 2\)), and \((\omega^{2} + 1)^{2026} = \omega^{2026} = \omega\). The roots are the reciprocals \(\omega\) and \(\omega^{2}\), so the equation is \(x^{2} + x + 1 = 0\).

(Q) With the odd power 2027 a minus sign survives: \((-\omega^{2})^{2027} = -\omega\) and \((-\omega)^{2027} = -\omega^{2}\). The roots are \(-\omega^{2}\) and \(-\omega\), with sum 1 and product 1, so the equation is \(x^{2} - x + 1 = 0\).

(R) The roots of \(x^{2} - x + 1 = 0\) are \(-\omega\) and \(-\omega^{2}\), so \(\gamma - 1 = \omega^{2}\) and \(\delta - 1 = \omega\). The sum is \(\frac{1}{\omega^{4052}} + \frac{1}{\omega^{2026}} = \omega + \omega^{2} = -1\).

(S) Put \(a = p + 1\), \(b = r + 1\). They satisfy \(y^{2} - y - 1 = 0\), so \(a + b = 1\) and \(ab = -1\):

\[ \begin{aligned} \frac{1}{a^{3}} + \frac{1}{b^{3}} &= \frac{(a + b)^{3} - 3ab(a + b)}{(ab)^{3}} \\ &= \frac{1 + 3}{-1} = -4 \end{aligned} \]

Only the third option has all four pairings. The first swaps R and S, the fourth also gets P and Q wrong, and the second gets P and Q wrong.