Question

If `a+b+c !=0` and `a^3+b^3+c^3-3abc=0`, then what will be the relation between `a`, `b`. and `c`?

Answer 1

See below.

Explanation:

Considering that

`(a+b+c)^3-(a^3 + b^3 + c^3 - 3 a b c) = 3(a+b+c)(ab+bc+ac)`

then if `(a+b+c) ne 0` we have

`(a+b+c)^2= 3(ab+bc+ac)` and finally

`a+b+c = pm sqrt3 sqrt(ab+bc+ac)`

NOTE:

`0=(a+b+c)^2- 3(ab+bc+ac) = a^2+b^2+c^2-(ab+ac+bc) = 1/2((a-b)^2+(b-c)^2+(a-c)^2)=0`

Answer 2

`a=b=c`

Explanation:

As it given that,
`a^3+b^3+c^3-3abc=0` then
`(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
since,
`a+b+c` not equal to `0` so,
`a^2+b^2+c^2-ab-bc-ca=0`
multiplying both sides by `2`
`2a^2+2b^2+2c^2-2ab-2bc-2ca=0` or
`(a-b)^2+(b-c)^2+(c-a)^2=0`
to satisfy the equation,
`(a-b)^2=0,(b-c)^2=0` and#(c-a)^2=0 rArr a=b,b=c,c=a#
hence
`a=b=c` proved