Question

25. If a+b+c = 6,a^2 + b^2 + c^2 =12 ,find the value of a^3 + b^3 + c^3 - 3abc.​

Answer 1

Answer:

a+b+c=6,a

2

+b

2

+c

2

=14,a

3

+b

3

+c

3

=36

(a+b+c)=a

2

+b

2

+c

2

+2(ab+bc+ca)

⟹2(ab+bc+ca)=(6)

2

−14

⟹2(ab+bc+ca)=(6)

2

−14

=36−14=22

ab+bc+ca=11

Now ,we know a

3

+b

3

+c

3

−3abc=(a+b+c)(a

2

+b

2

+c

2

−(ab+bc+ca))

⟹36−3abc=6(14−11)

⟹36−3abc=6×3=18

⟹3abc=18

⟹abc=6