Question

divide the following expression - 108 a³ b³ c by 18 abc​

Answer 1

Step-by-step explanation:

Given that( a+b+c)=6, a^2+b^2+c^2 =14, a^3+b^3+c^3 =36

There is an identity

a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) ………(1)

Taking (a+b+c) =6 ,squarreing both sides

(a+b+c)^2 = 36

a^2+b^2+c^2+2ab+2bc+2ca =36

14 +2ab+2bc+2ca=36

2(ab +bc+ca)= 36–14=22

ab+bc+ ca= 22/2

ab+bc+ca= 11,………..(2)

Putting the value of ab+bc+ ca =11 ,a^2+b^2+c^2 , a+b+c,a^3+b^3+c^3 in eq(1)

36–3abc=6(14–11)

36–3abc=6×3

-3abc =18–36

-3abc=-18

abc= 18/3=6.