Suppose that for some a,b,c we have \(a+b+c = 1\), \(ab+ac+bc = abc = -4\). What is \(a^3+b^3+c^3?\)
+128475 a + b + c = 1 cube both sides
a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3 = 1
(a^3 + b^3 + c^3) + 3a(ab) + 3a ( ac) + 3b (ab) + 3c (ac) + 3b(bc) + 3c(bc) + 6abc = 1
(a^3 + b^3 + c^3) + 3a ( ab + ac) + 3b( ab + bc) + 3c(ac + bc) + 6abc = 1
(a^3 + b^3 + c^3) + 3a ( -4 - bc) + 3b ( -4 - ac) + 3c ( -4 - ab) + 6abc = 1
(a^3 + b^3 + c^3) - 12a - 3abc - 12b - 3abc - 12c - 3abc + 6abc = 1
(a^3 + b^3 + c^3) - 12 ( a + b + c) - 9abc + 6abc = 1
(a^3 + b^3 + c^3) - 12 ( 1) - 3abc = 1
(a^3 + b^3 + c^3) - 12 - 3(-4) = 1
(a^3 + b^3 + c^3) = 1
