Ta có:
`a^3 + b^3 + c^3 = 3abc`
`<=> a^3 + b^3 + c^3 - 3abc = 0`
`<=> (a^3 + ab^2 + ac^2 - ab^2 - abc - a^2c) + (a^2b + b^3 + bc^2 - ab^2 - cb^2 - abc) + (a^2c + b^2c + c^3 - abc - bc^2 - ac^2) = 0`
`<=> a (a^2 + b^2 + c^2 - ab - bc - ac) + b (a^2 + b^2 + c^2 - ab - bc - ac) + c (a^2 + b^2 + c^2 - ab - bc - ac) = 0`
`<=> (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ac) = 0`
`<=> a + b + c = 0` hoặc `a^2 + b^2 + c^2 - ab - bc - ac = 0`
`+) a + b + c = 0`
`+) a^2 + b^2 + c^2 - ab - bc - ac = 0`
`<=> 2 (a^2 + b^2 + c^2 - ab -bc-ac) = 0`
`<=> 2a^2+ 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0`
`<=> (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) = 0`
`<=> (a-b)^2 + (b-c)^2 + (a-c)^2 = 0`
`\forall a;b;c` ta có :
`(a-b)^2 \ge 0`
`(b-c)^2 \ge 0`
`(a-c)^2 \ge 0`
`=> (a-b)^2 + (b-c)^2 + (a-c)^2 \ge 0`
Dấu `=` xảy ra `<=> {((a-b)^2 = 0 ),((b-c)^2 = 0 ),((a-c)^2 = 0):}`
`<=> {(a-b=0),(b-c=0),(a-c=0):}`
`<=> {(a = b ),(b = c ),(a = c):}`
`<=> a = b = c`
Vậy với `a^3 + b^3 + c^3 = 3abc ` thì `a+b+c=0` hoặc `a=b=c`.
