x² + 5x = 1 => x² + 5x - 1 = 0
=>
α+β = -5
αβ = -1

(i)
α² + 5α = 1
α² = 1 - 5α
α³ = α - 5α² = α - 5( 1 - 5α)
= α - 5 + 25α
= 26α - 5 (shown)


(ii)
since α³ = 26α - 5, then
β³ = 26β - 5 (since α & β are roots. )

sum of roots:
α³ +β³ = 26α + 26β - 10
= 26(α+β) - 10
= 26(-5) - 10
= -140

product of roots:
α³ x β³ = (αβ)³ = (-1)³ = -1

thus,
equation is:
x² + 140x - 1 = 0

Good job Snell! I did not think of the logic for part (ii) by using the part (i) analogy for the beta root. I would have just written it the normal factorisation way.