Question

A) If a +b+c=6, a2 + b2 +02= 14 and a3+b3+c3=36 then prove
that abc=6.​

Answer 1

Answer:

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

(6)^2=14+2(ab+bc+ca)

36-14=2(ab+bc+ca)

22/2=ab+bc+ca

ab+bc+ca=11

a^3+b^3+c^3-3abc

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

=(6)(14-11)

=(6)(3)

=18

Since, a^3+b^3+c^3-3abc=18

36-3abc=18

-3abc=18-36

-3abc=-18

abc=6

hope it helps you.....