Nốt bài 70 nhé :"v
Ta có: \( a^3 + b^3 + c^3 = 3abc \)
\(\Leftrightarrow a^3 + b^3 + c^3 - 3abc = 0 \)
\(\Leftrightarrow\)\( a^3 + b^3 + 3a^2b + 3ab^2 - 3a^2b - 3ab^2 + c^3 - 3abc = 0 \)
\(\Leftrightarrow(a+b)^3 - 3a^2b - 3ab^2 + c^3 - 3abc = 0 \)
\(\Leftrightarrow[(a+b)^3+c^3]-3ab(a+b+c)=0\)
\(\Leftrightarrow(a+b+c)[(a+b)^2 - c(a+b) + c^2] – 3ab(a+b+c) = 0 \)
\(\Leftrightarrow(a+b+c)(a^2+ 2ab + b^2 - ac – bc + c^2 - 3ab) = 0 \)
\(\Leftrightarrow(a+b+c)(a^2+b^2+c^2-bc-ab-ca)=0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)
Vì \(a,b,c\) là số dương nên \(a+b+c>0\)
\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow a-b=b-c=c-a\)
\(\Rightarrow a=b=c\) ( Đpcm )