Ta có: \(a=b=c\Rightarrow\hept{\begin{cases}a^3=abc\\a^3=b^3=c^3\end{cases}}\)

Vì \(a^3=b^3=c^3\Rightarrow a^3+b^3+c^3=3a^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

\(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3+c^3=0\)

\(\Leftrightarrow a^3-3abc+b^3+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

\(a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

hình như sai đề phải ko các bạn

\(a^3+b^3+c^3=3abc\) 

<=>   \(a^3+b^3+c^3-3abc=0\)

<=>    \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>    \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

  Xét:     \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>    \(2a^{ 2}+2b^2+2c^2-2ab-2bc-2ca=0\)

<=>     \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>    \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\) <=>  \(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)<=>   \(a=b=c\)

=>  đpcm

a³ + b³ + c³ = 3abc 
<=> a³ + b³ + c³ - 3abc = 0 
<=> a³ + b³ + 3a²b + 3ab² - 3a²b - 3ab² + c³ - 3abc = 0 
<=> (a+b)³ - 3a²b - 3ab² + c³ - 3abc = 0 
<=> [(a+b)³ + c³] – 3ab(a + b + c) = 0 
<=> (a+b+c)[(a+b)² - c(a+b) + c²] – 3ab(a+b+c) = 0 
<=> (a+b+c)(a² + 2ab + b² - ac – bc + c² - 3ab) = 0 
<=> (a+b+c)(a² + b² + c² - bc – ab – ca) = 0 
<=>{a + b +c = 0, a;b;c là các số dương => a = b = c 
hoặc {a² + b² + c² - bc – ab – ca = 0 
<=> 2a² + 2b² + 2c² - 2bc – 2ab – 2ca = 0 
<=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0 
<=> (a - b)² + (b - c)² + (c - a)² = 0 
mà (a - b)² ≥ 0 với mọi a;b 
(b - c)² ≥ 0 với mọi b;c 
(c - a)² ≥ 0 với mọi a;c 
nên ta có a - b = b - c = c - a 
=> a = b =c