Algebra

Note that $a^3 + b^3 + c^3 = \left(\left(a+b+c\right)^2-3ab-3bc-3ac\right)\left(a+b+c\right)+3abc$ (found it off the internet)

However, we can rewrite this into something easier: $\left(\left(a+b+c\right)^2-3(ab+bc+ac)\right)\left(a+b+c\right)+3abc$

Can you take it from here?

Builderboi has a great way to solve it but sometimes you don't have much of those formulas memorized.

Use polynomial construction.

a + b + c = 1

ab + ac + bc = -1

abc = -1

P(x) = x^3 - x^2 - x + 1

a, b, c are the roots of this polynomial. 

a^3 = a^2 + a - 1

b^3 = b^2 + b - 1

c^3 = c^2 + c - 1

a^3 + b^3 + c^3 = (a^2 + b^2 + c^2) + (a + b + c) - 3

(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 +  c^2 = 3

a^3 + b^3 + c^3 = 1.