Question

If a+b+c =6 and ab+bc+ac=11, find the value of a^3+b^3+c^3-3abc.​

Answer 1

Answer:

the solution is In the above.

Answer 2

Answer:

18

Step-by-step explanation:

(a+b+c)^2=ab + 2bc + 2ca

36=a^2 + b^2 + c^2 + 2(ab+bc+ca)

36=a^2 + b^2 + c^2 + 2(11)

36-22=a^2 + b^2 + c^2

14=a^2 + b^2 + c^2

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

=(6)(14-11)

=6*3=18