Answer: 8
Explanation:
EQ1: a + b + c = 2 --> b + c = 2 - a
EQ2: ab + bc + ac = -1 --> b + c = (-1 - bc)/a
EQ3: abc = -2 --> bc = -2/a
Set EQ1 = EQ2 and substitute bc using EQ3 to solve for "a":
Consider the solution a = 2 and plug it into EQ1 to solve for "b"
b + c = 2 - 2
b + c = 0
b = -c
Plug in a = 2, b = -c, and c = c into a³ + b³ + c³
2³ + (-c)³ + c³
= 8 - c³ + c³
= 8