CM a + b + c = 0
=> a + b = -c ; b + c = -a ; c+a a = -b
E = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=1\)
Như thế này :
\(a^3+b^3+c^3=3abc\)
=> (a+b)^3 - 3ab(a+b) - 3abc + c^3 = 0
=> ( a+ b +c )^3 - 3(a+b)c(a+b+c) - 3ab(a+b+c) = 0
=> \(\left(a+b+c\right)\left[\left(a+b+c\right)^2-3bc-3ac-3ab\right]=0\)
=> ( a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca ) = 0
=> 1/2 ( a + b + c )(2a^2 + 2b^2 + 2x^2 - 2ab - 2bc - 2 ca ) = 0
=> 1/2 (a+b+c) [ ( a- b)^2 + ( b - c)^2 + (c-a)^2] = 0
Bì ngoặc thứ hai luôn >= 0 => a + b + c = 0
hoặc a = b ; b =c = c=a => a = =b =c
