The equation x³ + 2x² + x + 7 = 0 has roots α, β, γ.
- (i)Use the relation x² = −7y to show that the equation 49y³ + 14y² - 27y + 7 = 0 has roots α/(βγ), β/(γα), γ/(αβ).[4]
- (ii)Show that α²/β²γ² + β²/γ²α² + γ²/α²β² = 58/49.[3]
- (iii)Find the exact value of α³/β³γ³ + β³/γ³α³ + γ³/α³β³.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [M1] Uses given substitution and eliminates radical.
- [A1] AG
- [M1] Uses αβγ = -7.
- [A1] AG
- [B1] States sum of roots and α'β'+α'γ'+β'γ'.
- [M1 A1] Uses α'^2 + β'^2 + γ'^2 = (α'+β'+γ')^2 - 2(α'β'+α'γ'+β'γ'). AG
- [M1] Uses 49α'^3 = -14α'^2 + 27α' - 7.
- [A1]
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