Question

If a+b+c=6 and ab+bc+ac=11, find the value of a3+b3+c3

-3abc.

Answer 1

Hello friend

We have an algebraic identity

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

It is given that a+b+c=6 and ab+bc+ac=11

6²=a²+b²+c²+2(11)

36-22=a²+b²+c²

So a²+b²+c²=14
____________________________

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-(ab+bc+ca))

a³+b³+c³-3abc=6×(14-11)

=6×3

=18

Hope it helps!