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`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Có: \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3-3abc=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
2)Có: \(a+b-c=0\)
\(\Leftrightarrow a+b=c\)
\(\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3\)
\(\Leftrightarrow a^3+b^3+3abc=c^3\)
\(\Leftrightarrow a^3+b^3-c^3=-3abc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a +b +c=0
⇔\(\left(a+b+c\right)^3\)
⇔\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)
⇔\(a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)
⇔ \(a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)=3abc\)
Vì a+b+c= 0
⇒\(a^3+b^3+c^3=3abc\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) (chuyển vế qua)
\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Do VP >=0 với mọi a, b, c. Nên để đăng thức xảy ra thì a = b = c
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(-c\right)\)
\(\Rightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
(Nhớ k cho mình với nhá!)
Ta có :(a+b+c)3=a3+b3+c3+3a2b+3a2c+3b2c+3b2a+3c2a+3c2b+6abc
(a+b+c)3=a3+b3+c3+(3a2b+3a2b+3abc)+(3b2c+3b2a+3abc)+(3c2a+3c2b+3abc)-3abc
(a+b+c)3=a3+b3+c3+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)-3abc
(a+b+c)3=a3+b3+c3+3(a+b+c)(ab+bc+ac)-3abc
thay a+b+c=0 ta được
03=a3+b3+c3+3.0(ab+bc+ac)-3abc
0=a3+b3+c3-3abc
=>a3+b3+c3=3abc
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Từ $a+b+c=0\Rightarrow a+b=-c$
Theo HĐT đáng nhớ:
\((a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)\)
\(\Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)=(-c)^3-3ab(-c)=-c^3+3abc\)
\(\Rightarrow a^3+b^3+c^3=-c^3+3abc+c^3=3abc\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có a+b+c=0
=>a+b=-c
ta có a^3 +b^3+c^3
=(a+b)(a^2-ab+b^2)+c^3
=-c(a^2+b^2-ab)+c^3
=-c[(a+b)^2-2ab-ab]+c^3
= -c[(-c)^2-3ab]+c^3
= (-c)^3+3abc+c^3
=3abc
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+ab^2+ac^2-a^2b-abc-ca^2+ba^2+b^3+bc^2-b^2a-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2=0\)
\(\Leftrightarrow a\left(a^2+b^2+c^2-ab-bc-ca\right)+b\left(a^2+b^2+c^2-ab-bc-ca\right)+c\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(luôn đúng)
\(\Rightarrowđpcm\)
mình có cách khác
a+b+c=0
=>a+b=-c
=>a+b=3abc/-3ab
=>(a+b).(-3ab)=3abc
=>(a+b).(a^2-ab+b^2-a^2-2ab-b^2)=3abc
=>(a+b)(a^2-ab+b^2)-(a+b).(a^2+2ab+b^2)=3abc
=>a^3+b^3-(a+b)^3=3abc
mà a+b=-c=>
a^3+b^3-(-c)^3=3abc
=>a^3+b^3+c^3=3abc
Ta có:a+b+c=0
=>a+b=-c
=>(a+b)^3=(-c)^3
a^3+3a^2b+3ab^2+b^3=(-c)^3
=>a^3+3a^2b+3ab^2+b^3+c^3=0
=>a^3+b^3+c^3+3ab(a+b)=0
=>a^3+b^3+c^3+3ab(-c)=0
=>a^3+b^3+c^3-3abc=0
=>a^3+b^3+c^3=3abc (đpcm)
Ta có:
a^3+b^3+c^3−3abc=(a+b)^3−3a^2b−3ab^2+c^3−3abc=[(a+b)^3+c^3]−3ab(a+b+c)
=(a+b+c)[(a+b)^2−(a+b)c+c^2]−3ab(a+b+c)=0
-> a^3+b^3+c^3=3abc