A Tricky Roots of Cubics Problem

Given that α, β, γ are roots of the equation x^3 + 2x + 1 = 0, we calculate α^5 + β^5 + γ^5, using some elegant algebraic techniques and shortcuts.
00:00 Simultaneous equations
02:46 Vieta's formulas
04:55 Sum of squares
06:14 Sum of cubes
11:48 Sum of fifth powers