Đáp án:
Giải thích các bước giải:
`A)a^3+b^3+c^3 =3abc`
`<=>a^3+b^3+c^3 -3abc=0`
`<=>(a^3+b^3+3a^2b+3ab^2)+c^3-3abc-3a^2b-3ab^2=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`<=>`\(\left[ \begin{array}{l}a+b+c=0(dpcm)\\a^2+b^2+c^2-ab-bc-ca=0\end{array} \right.\)
`B)a^3+b^3+c^3 =3abc`
`<=>a^3+b^3+c^3 -3abc=0`
`<=>(a^3+b^3+3a^2b+3ab^2)+c^3-3abc-3a^2b-3ab^2=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`<=>`\(\left[ \begin{array}{l}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{array} \right.\)
Vì `a^2+b^2+c^2-ab-bc-ca=0`
`<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`<=>`\(\left[ \begin{array}{l}a=b\\b=c\\c=a\end{array} \right.\)
`<=>a=b=c(dpcm)`
