Ta có: a3 + b3 + c3 = 3abc
<=> (a + b)(a2 - ab + b2) + c3 - 3abc = 0
<=> (a + b)3 - 3ab(a + b) + c3 - 3abc = 0
<=> (a + b + c)[(a + b)2 - (a + b)c + c2) - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> \(\orbr{\begin{cases}a+b+c=0\left(loại\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
<=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
<= > (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ac + a2) = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)<=> a = b = c
Khi đó: B = \(\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
ta có a3+b3+c3=3abc <=> a3+b3+c3-3abc=0
<=> (a+b)3-3ab(a+b)+c3-3abc=0
<=> (a+b+c)3-3(a+b)c(a+b+c)-3ab(a+b+c)=0
<=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0
<=> a2+b2+c2-ab-bc-ca=0 (vì a+b+c=0)
<=> (a-b)2+(b-c)2+(c-a)2=0
<=> a=b=c
khi đó \(B=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
