Question

if a+b+c=8 and ab+bc+ca=20 then find the value of a^3+b^3+c^3-3abc​

Answer 1

Answer:

32

Step-by-step explanation:

a+b+c=8

ab+bc+ca=20

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

(8)^2=a^2+b^2+c^2+2(ab+bc+ca)

64=a^2+b^2+c^2+2(20)

64=a^2+b^2+c^2+40

a^2+b^2+c^2=64-40

a^2+b^2+c^2=24

Now,a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

=(8)(24-20)

=8×4

=32