Question

Nếu a+b+c =0 hoặc a=b=c  thì a³+b³+c³=3(abc)

Answer 1

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

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

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

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)

Vì a+b+c=0 \(\hept{\begin{cases}a>0\\b>0\\c>0\end{cases}}\)

Do đó: \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow a=b=c}\)

Answer 2

Miyuki Misaki cm ngược rồi 
Ta có : a + b + c = 0 

<=> a + b = -c {...........}

<=> (a + b)³ = -c³

<=> a³ + b³ + 3ab(a + b) = -c³

<=> a³ + b³ + c³ = -3ab(a + b) 

<=> a³ + b³ + c³ = -3ab(-c) {vì a + b = -c}

<=>  a³ + b³ + c³ = 3abc