Question

if a+b+c = 6 and ab+bc+ac = 8 find a^3 +b^3+c^3-3abc​

Answer 1

Answer:

a3+b3+c3-3abc=a2+b2+c2-ab-bc-ca

=a2+b2+c2-(ab+bc+ca)

=a2+b2+c2-(8)

=(a+b+c)2-8

=(6)2-8

=12-8

=4

Answer 2

Answer:

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

Step-by-step explanation:

so

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

a^2+b^2+c^2=6^2-8=28

then

result is

6×(28-8)=120