Question

If a+b+c=6 and ab+bc+ca=11, find the value of a3+b3+c3-3abc.

Answer 1

a+b+c=6 -------------------------------(1)
Cubing both sides,
(a+b+c)³=6³
or, (a+b)³+3(a+b)²c+3(a+b)c²+c³=216
or, a³+3a²b+3ab²+b³+3c(a²+2ab+b²)+3ac²+3bc²+c³=216
or, a³+b³+c³+3a²b+3ab²+3a²c+3b²c+3ac²+3bc²+6abc=216
or, a³+b³+c³+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc=216
or, a³+b³+c³+3ab(6-c)+3ac(6-b)+3bc(6-a)+6abc=216 [using (1)]
or, a³+b³+c³+18ab-3abc+18ac-3abc+18bc--3abc+6abc=216
or, a³+b³+c³+18(ab+bc+ca)-9abc+6abc=216
or, a³+b³+c³+18×11-3abc=216 [∵, ab+bc+ca=11]
or, a³+b³+c³-3abc=216-198
or, a³+b³+c³-3abc=18