Question

if a+b+c=6 , a²+b²+c²=14 and a³+b³+c³=36 then find the value of (abc) ?

Answer 1

(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