If α and β are the zeroes of $x^2-4x-5$, the value of α³ + β³ is:

$\alpha +\beta =4$ and $\alpha \beta =-5.{\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=64-3(-5)(4)=64+60=124$. (Check with the zeroes 5 and −1: $125+(-1)=124$.) 4 keeps the wrong sign on the $-3\alpha \beta (\alpha +\beta )$ term; 64 stops at $(\alpha +\beta {)}^3$; 79 drops the factor $(\alpha +\beta )$ and computes $64+15$.