+ \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^2-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc-3ab\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) ( do \(a+b+c\ne0\) )\
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Ta có : \(a^3+b^3+c^3=3\cdot abc\)
\(\Rightarrow a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)
\(=>a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Lại có : a,c,b là các số dương=>a+b+c\(\ne0\)
Mà a+b+c=0 , \(a^2+b^2+c^2-ab-ac-bc=0\)
=>a=b=c(=0)(đpcm)
