a+b+c=0

=>(a+b+c)³=0

=>a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+....mk phải ăn cơm rồi

Xét \(a^3+b^3+c^3-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Mà \(a+b+c=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\) 

 thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

a^3+b^3+c^3-3abc=0 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0 

<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)... 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0 

luôn đúng do a+b+c=0

1 ) Ta có :

\(a+b-c=0\Leftrightarrow a+b=c\Leftrightarrow\left(a+b\right)^3=c^3\)

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

\(\Rightarrow a^3+b^3-c^3=a^3+b^3-3a^2b-3b^2a-b^3\)

\(\Rightarrow a^3+b^3-c^3=-3a^2b-3b^2a\)

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

\(\Rightarrow a^3+b^3-c^3=-3abc\left(đpcm\right)\)

2 ) Ta có :

\(a-b+c=0\Leftrightarrow c=b-a\Leftrightarrow c^3=\left(b-a\right)^3\)

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

\(\Rightarrow a^3-b^3+c^3=a^3-b^3+b^3-3a^2b+3b^2a-a^3\)

\(\Rightarrow a^3-b^3+c^3=-3a^2b+3b^2a\)

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

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

\(\Rightarrow a^3-b^3+c^3=3abc\left(đpcm\right)\)

1 ) Bổ sung dấu \(\Rightarrow\) thứ 2 :

\(\Rightarrow...=a^3+b^3-a^3-3a^2b-3b^2a-b^3\)

Ta có : \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

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

Lại có : \(a^3+b^3+c^3\)

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

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

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

\(=-3ab.\left(-c\right)\)

\(=3abc\left(đpcm\right)\)

:D

a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 -ab - bc - ac) + 3abc

hiểu rồi chứ?

Ta có : a + b + c = 0

\(\Rightarrow\)a + b = - c

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\\ \Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\\ \Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\\ \Rightarrow a^3+b^3+c^3=-3ab.\left(-c\right)\\ \Rightarrow a^3+b^3+c^3=3ab\left(đpcm\right)\)

ta có:a+b=(-c)

(a+b)^3=(-c)^3

a^2+3a^2b+3ab^2+b^3=(-c)^3

a^3+b^3+c^3= -3a^2b+3ab^2

a^3+b^3+c^3= -3ab(a+b)

a^3+b^3+c^3= -3ab(-c)

a^3+b^3+c^3=3abc

Ta có: a+b+c = 0

=> a+b = -c

a^3+b^3 +c^3 = (a+b)^3 - 3a^2.b - 3ab^2 +c^3

= (-c)^3 - 3ab(a+b) +c^3

= (-c)^3 +c^3 - 3ab.(-c) = -3ab(-c) = 3abc (đpcm)

Ta có: a³ + b³ + c³ - 3abc

= (a³ + b³) + c³ - 3abc

= (a + b)³ - 3ab(a + b) + c³ - 3abc

= [(a + b)³ + c³ ] - [3ab(a + b) - 3abc]

= (a + b + c)[(a + b)² - (a + b)c + c² ] - 3ab(a + b + c)

= (a + b + c)(a² + b² + c² - ab - bc - ca)

mà a + b + c = 0

=> a³ + b³ + c³ - 3abc = 0

=> a³ + b³ + c³ = 3abc (đpcm)

Chúc bạn học tốt!

a + b + c = 0

⇔ a + b = -c

⇔ (a + b)³ = -c³

⇔ a³ + b³ + 3ab(a + b) = -c³

⇔ a³ + b³ - 3abc = -c³

⇔ a³ + b³ + c³ = 3abc