Ta có:
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\) (1)
Mà \(a+b+c=0\)
\(\left(1\right)\Rightarrow\frac{1}{2}.0.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vậy: nếu \(a+b+c=0\) thì \(a^3+b^3+c^3-3abc=0\)
Chúc bạn học tốt và tíck cho mìk vs nha bùi thị thu hương!
ta có\(a+b+c=0\)
\(a+b=-c\)
\(b+c=-a\)
\(a+c=-b\)
ta lại có VT: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(a+c\right)-3abc\)
\(=\left(a+b+c\right)^3-3\left(-c\right)\left(-a\right)\left(-b\right)-3abc\)
\(=\left(a+b+c\right)^3+3abc-3abc\)
\(=0=VP\)
nghe nhe',bai nay de thui ma.
ta xet ve trai a^3+b^3+c^3=
[(a+b)(a^2-ab+b^2)]+c^3 dung ko.(1)
ma ta co theo gia thiet a+b+c=0 suy ra c= - (a+b)suy ra
c^3= -(a+b)^3
thay vao`(1) ta co [(a+b)(a^2-ab+b^2)] - (a+b)^3
(lay nhan tu chung ta co)=(a+b)[a^2-ab+b^2-(a+b)^2]
(phan h (a+b)^2) =(a+b)[a^2-ab+b^2-(a^2+2ab+b^2)]
=(a+b)(a^2-ab+b^2-a^2-2ab-b^2)
=(a+b).(-3ab)
= -(a+b).3ab (2)
theo gia thiet ta co a+b+c=0 suy ra c= -(a+b)
thay vao(2) ta dc
=3abc
vay la xong
ket luan ve trai bang ve phai
neu chua hieu thi chat vao nick i_hate_i_love_i anh se giai thix cho.:D
Ta có :
\(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)[\left(a+b\right)^2-ac-bc+c^2]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)\)
\(=0\left(a^2+b+c^2-ab-ac-ab\right)=0\left(đpcm\right)\)
Ta có : a3+b3+c3=3abc⇔a3+b3+c3−3abc=0a3+b3+c3=3abc⇔a3+b3+c3−3abc=0
⇔(a+b)3+c3−3ab(a+b)−3abc=0⇔(a+b)3+c3−3ab(a+b)−3abc=0
⇔(a+b+c)(a2+b2+2ab−bc−ac)−3ab(a+b+c)=0⇔(a+b+c)(a2+b2+2ab−bc−ac)−3ab(a+b+c)=0
⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ac)=0
⇔a+b+c2[(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)]=0⇔a+b+c2[(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)]=0
⇔a+b+c2[(a−b)2+(b−c)2+(c−a)2]=0⇔a+b+c2[(a−b)2+(b−c)2+(c−a)2]=0
⇔[a+b+c=0(a−b)2+(b−c)2+(c−a)2=0⇔[a+b+c=0(a−b)2+(b−c)2+(c−a)2=0
⇔[a+b+c=0a=b=c
