a + b + c = 6 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 = 216
(a^3 + b^3 + c^3) + (6abc) + 3a^2b + 3ab^2 + 3a^2c + 3ac^2 + 3b^2c + 3bc^2 = 216
(a^3 + b^3 + c^3) + 6( -12) + 3ab (a + b) + 3ac ( a + c) + 3bc ( b + c) = 216
(a^3 + b^3 + c^3) + 3ab ( 6 - c) + 3ac ( 6 - b) + 3bc ( 6 - a) = 216 + 6(12)
(a^3 + b^3 + c^3) + 18ab - 3abc + 18ac - 3abc + 18bc - 3abc = 288
(a^3 + b^3 + c^3) + 18 ( ab + ac + bc) - 9abc = 288
(a^3 + b^3 + c^3) + 18 (5) - 9 ( -12) = 288
(a^3 + b^3 + c^3) + 90 + 108 = 288
(a^3 + b^3 + c^3) + 198 = 288
a^3 + b^3 + c^3 = 288 - 198 = 90
Reposted here https://web2.0calc.com/questions/update-i-m-still-confused
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