Question

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

Answer 1

a*3+b*3+c*3-3abc
=a*3+b*3+c*3-3abc +3(a+b+c)(ab+bc+ac)-3(a+b+c)(ab+bc+ac)
=(a+b+c)*3 - 3(a+b+c)(ab+bc+ac)
=6*3 - 3×6×11
=216-198
=18

Answer 2

First see the expansion of (a+b+c)³

(a+b+c)³

= [(a+b)+c]³

= (a+b)³+3(a+b)²c+3(a+b)c²+c³

(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c²+c³

=a³+b³+c³+3a²b+3a²c+3ab²+3b²c+3ac²+3bc²+6abc

=(a³+b³+c³)+(3a²b+3a²c+3abc)+(3ab²+3b²c+3abc)+(3ac²+3bc²+3abc)−3abc


=(a³+b³+c³)+3a(ab+ac+bc)+ 3b(ab+bc+ac) +3c(ac+bc+ab)−3abc

=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc

OR
(a+b+c)³=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc

Now, we know the expansion.
Given that a+b+c=6 and ab+bc+ca=11

(a+b+c)³=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc

(6)³=(a³+b³+c³)+3(6)(11)−3abc

216=(a³+b³+c³)+198-3abc

(a³+b³+c³)-3abc=216-198

(a³+b³+c³)-3abc= 18

Hence your answer is 18.