The equation x³ + x − 1 = 0 has roots α, β, γ. Use the relation x = √y to show that the equation y³ + 2y² + y − 1 = 0 has roots α², β², γ².

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

1. [1] (a) Substitute x = √y into x³ + x - 1 = 0: (√y)³ + √y - 1 = 0. Rearrange to y√y + √y = 1 ⇒ √y(y + 1) = 1
2. [1] (a) Square both sides: y(y + 1)² = 1 ⇒ y(y² + 2y + 1) = 1 ⇒ y³ + 2y² + y - 1 = 0 (AG)
3. [1] (i) For x³ + x - 1 = 0, S₁ = 0, S₂ = -2 (from Newton sums: S₂ + S₁p₁ + 2p₂ = 0, where p₁=0, p₂=1)
4. [2] (i) For y³ + 2y² + y - 1 = 0, S'₄ + 2S'₂ + S'₁ - 3 = 0, or use S₄ = α⁴ + β⁴ + γ⁴. Use Newton sums for x³+x-1=0: S_n + p₁S_(n-1) + p₂S_(n-2) + p₃S_(n-3) = 0. For n=4, S₄ + 0 S₃ + 1 S₂ - 1 S₁ = 0 ⇒ S₄ + S₂ - S₁ = 0. Given S₁ = 0 and S₂ = -2, S₄ + (-2) - 0 = 0 ⇒ S₄ = 2 (AG)
5. [2] (ii) Using Newton sums for x³+x-1=0: S_n + S_(n-2) - S_(n-3) = 0. For n=6, S₆ + S₄ - S₃ = 0. S₃ = S₂ - S₀ + p₃ = -2 - 3 + 1 = -4. (Here, S₃ + p₁S₂ + p₂S₁ + p₃S₀ = 0 => S₃ + 0S₂ + 1S₁ - 1(3) = 0 => S₃ - 3 = 0, so S₃=3. Re-evaluate S₃: p₁=0, p₂=1, p₃=-1. S₃+p₁S₂+p₂S₁+3p₃=0 => S₃+0+1(0)+3(-1)=0 => S₃=3.) So, S₆ + 2 - 3 = 0 ⇒ S₆ = 1
6. [2] (ii) For n=8, S₈ + S₆ - S₅ = 0. Need S₅. S₅ + S₃ - S₂ = 0 ⇒ S₅ + 3 - (-2) = 0 ⇒ S₅ = -5. So, S₈ + 1 - (-5) = 0 ⇒ S₈ = -6