If a + b + c = 6, $a^2+b^2+c^2=32$, and $a^3+b^3+c^3=189$, then the value of abc - 3 is:

2 3 1 0

0

a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca) a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - (ab + bc + ca)) If a + b + c = 6 $a^2+b^2+c^2=32$ $a^3+b^3+c^3=189$ 32 = (6)2 - 2(ab + bc + ca) 32 = 36 - 2(ab + bc + ca) (ab + bc + ca) = 2 189 - 3abc = (6) (32 - (2)) 3abc = 9 abc = 3 abc - 3 = 0