Question 13

For the roots α=ω, β=ω² of x²+x+1=0, I compute α+1 = -ω², so (α+1)^2026 = ω^4052 = ω² since 4052 mod 3 = 2, giving 1/(α+1)^2026 = ω. Similarly β+1 = -ω^4 = -ω. Working out (β+1)^2026 gives ω^2026 = ω (since 2026 mod 3 = 1), so 1/(β+1)^2026 = ω². These two values ω and ω² are exactly the roots of x²+x+1=0, matching (P) to option (1). For (Q), I compute (α+1)^2027 and (β+1)^2027 similarly, getting -ω^4054 = -ω and -ω^2027 = -ω², so their reciprocals are -ω² and -ω. These sum to 1 and multiply to 1, satisfying x²-x+1=0, matching (Q) to option (2). For (R) with γ,δ roots of x²-x+1=0, I identify γ=-ω, δ=-ω² and compute γ-1=ω², δ-1=ω. Raising to the 2026 power and taking reciprocals gives ω and ω², which sum to -1, matching (R) to option (4). Now for (S), with p,r roots of x²+x-1=0, I'm substituting y=1/(p+1) to express p in terms of y and plug it back into the original equation to find what quadratic y satisfies. **Solution sketch** Let ω be a primitive cube root of unity, so α = ω, β = ω². **(P)** α+1 = −ω², β+1 = −ω. Then (α+1)^2026 = ω^4052 = ω², (β+1)^2026 = ω^2026 = ω. Reciprocals: ω and ω² → equation x² + x + 1 = 0. **(P)→(1)** **(Q)** (α+1)^2027 = −ω^4054 = −ω, (β+1)^2027 = −ω^2027 = −ω². Reciprocals: −ω², −ω. Sum = 1, product = 1 → x² − x + 1 = 0. **(Q)→(2)** **(R)** Roots of x² − x + 1 = 0 are −ω, −ω². γ−1 = ω², δ−1 = ω. (γ−1)^2026 = ω², (δ−1)^2026 = ω. Sum of reciprocals = ω + ω² = −1. **(R)→(4)** **(S)** Let y = 1/(p+1) ⇒ p = 1/y − 1. Substituting in x² + x − 1 = 0 gives y² + y − 1 = 0, so y₁ + y₂ = −1, y₁y₂ = −1. y₁³ + y₂³ = (−1)³ − 3(−1)(−1) = −4. **(S)→(5)** Hence (P)→(1), (Q)→(2), (R)→(4), (S)→(5): option **C**.

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**Calculating polynomial roots** I'm working with the roots of the equation x^2 - x + 1. For the powers, I found that γ^3 = -1 and γ^4 = -γ, while δ^4 = -δ. The sum results in -(γ + δ), leading to -1. However, I realized R should be 4, not -4. Options A and D indicate R5, and B and C suggest R4. The values of P and Q are 1 and 2, which leans toward option C if S points to 5. Now, I need to calculate S.

Since α and β are primitive cube roots of unity, I note α+1 = -β and β+1 = -α, so I can rewrite 1/(α+1)^n and 1/(β+1)^n in terms of powers of α and β, then start building the sum of roots for the new quadratic equation. Since αβ=1, the product of the new roots simplifies to exactly 1 regardless of n. For the sum, I compute α^n+β^n for n=2026 using ω^n+ω^{2n}, finding that 2026 mod 3 = 1, so this reduces to ω+ω² = -1, giving a sum of new roots equal to -1, so the resulting quadratic is again x²+x+1=0 — meaning the transformation cycles back to the original equation. Repeating this for n=2027, I find 2027 mod 3 = 2, and again ω^n+ω^{2n} = -1, but the sign flips due to (-1)^{2027}, so the sum becomes 1, giving x²-x+1=0. This matches the pattern needed for options A and C, so now I move to checking R. I need γ-δ where γ,δ satisfy x²-x+1=0, using γ+δ=1 and γδ=1 to derive the relationship. So γ-1=γ² and δ-1=δ², which means I need 1/γ^4052 + 1/δ^4052. Since γ,δ are related to primitive 6th roots of unity (γ=e^{iπ/3}, δ=e^{-iπ/3}), this simplifies using their periodicity. Reducing 4052 mod 6 gives remainder 2, so γ^4052=γ² and δ^4052=δ². Then 1/γ²+1/δ² = (γ²+δ²)/(γδ)². Since γδ=1 (product of roots) and γ²+δ²=(γ+δ)²-2γδ=1-2=-1, the answer comes out to -1. Now I'm checking case S with p,r roots of x²+x-1=0, so p+r=-1 and pr=-1. I need 1/(p+1)³+1/(r+1)³, and I'm substituting u=p+1 to transform the original equation into u²-u-1=0, meaning u²=u+1, with r+1 satisfying the same relation. Computing u+v=1 and uv=-1, then u³+v³=(u+v)³-3uv(u+v)=1+3=4, while (uv)³=-1, giving the sum as 4/(-1)=-4. So S maps to option (5), and combining everything: P→1, Q→2, R→4, S→5, which matches option C. Looking at this problem, I need to analyze each part using properties of roots of unity. **Part P & Q:** For x²+x+1=0, roots α,β are primitive cube roots of unity with α+β=-1, αβ=1, so α+1=-β, β+1=-α. Using ω^n patterns (period 3) for n=2026 (≡1 mod 3) and n=2027 (≡2 mod 3), computing sum and product of new roots: - **P** (n=2026): Sum = -1, Product = 1 → x²+x+1=0 → **(1)** - **Q** (n=2027): Sum = 1, Product = 1 → x²-x+1=0 → **(2)** **Part R:** For x²-x+1=0, roots γ,δ are primitive 6th roots of unity, γ-1=γ², δ-1=δ² (from equation), γδ=1, γ+δ=1. Since γ^6=1, and 2026 mod 6 = 2·2026 mod 6... computing (γ-1)^2026 = γ^4052, and 4052 mod 6 = 2: 1/γ² + 1/δ² = (γ²+δ²)/(γδ)² = (1-2)/1 = -1 → **(4)** **Part S:** For x²+x-1=0, roots p,r with p+r=-1, pr=-1. Let u=p+1, v=r+1; these satisfy u²-u-1=0, giving u+v=1, uv=-1. 1/u³+1/v³ = (u³+v³)/(uv)³ = [(u+v)³-3uv(u+v)]/(uv)³ = [1+3]/(-1) = -4 → **(5)** **Final Matching:** P→(1), Q→(2), R→(4), S→(5) This corresponds to option **(C)**. ```json