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:
Answer & explanation
Correct answer: option 4
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