Question

1. Cho (a-b)²+(b-c)²+(c-a)²-3(a-b)(b-c_(c-a)=(a+b-2c)²+(b+c-2a)²+(c+a-2b)2
Chứng minh rằng a=b=c
2.Cho a+b+c=0. Chứng minh rằng : a³+b³+c³=3abc
3. Áp dụng chứng minh : (x-y)³+(y-z)³+(z-x)³= 3(x-y)(y-z)(z-x)
4.Cho a,b,c là các số dương và a³+b³+c³=3abc
Chứng minh rằng :a=b=c

Answer 1

Bài 2:

\(a^3+b^3+c^3-3bac\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3bac\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]\)

=0

=>\(a^3+b^3+c^3=3bac\)

Bài 4:

\(a^3+b^3+c^3=3bac\)

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

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3bac=0\)

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ba\left(a+b+c\right)=0\)

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

=>\(a^2+b^2+c^2-ab-bc-ac=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

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

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

=>a=b=c