Question

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

Answer 1

hope this helps..........

Answer 2

(a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc]
               =(a3+b3+c3)+3(a+b+c)(ab+ac+bc)−3abc
(a+b+c)3-3(a+b+c)(ab+ac+bc)=(a3+b3+c3)−3abc
⇒a3+b3+c3-3abc=6∧3-3*6*11
                            =216-198
                            =18