cho a^3 +b^3+c^3=3abc .Cm a+b+c = 0
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a + b + c = 0
\(\Leftrightarrow\)(a + b + c)3 = 0
\(\Leftrightarrow\)a3 + b3 + c3 + 3(a2b + ab2 + a2c + ac2 + b2c + bc2 + 2abc) = 0
\(\Leftrightarrow\)a3 + b3 + c3 + 3[ab(a + b + c) + ac(a + b + c) + bc(b + c)] = 0
\(\Leftrightarrow\)a3 + b3 + c3 + 3bc(b + c) = 0
\(\Leftrightarrow\)a3 + b3 + c3 = -3bc(b + c)
mà a + b + c = 0
\(\Rightarrow\)b + c = -a
Ta có: a3 + b3 + c3 = -3bc(b + c)
\(\Leftrightarrow\)a3 + b3 + c3 = -3bc(-a)
\(\Leftrightarrow\)a3 + b3 + c3 = 3abc (đpcm)
Ta có \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Rightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)
Mà \(a+b=-c\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a + b + c = 0 => a = -b - c
VT = \(a^3+b^3+c^3=\left(-b-c\right)^3+b^3+c^3\)
= \(\left(-b\right)^3-3\left(-b\right)^2c-3bc^2-c^3+b^3+c^3\)
= \(-3b^2c-3cb^2\)
= \(3bc\left(-b-c\right)\)
= \(3abc=VP\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng hằng đẳng thức:\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}\left(1\right)}\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\left(2\right)\)
Lấy (1) thay vào (2) ta được:
\(\Leftrightarrow a^3+b^3+c^3+3.\left(-c\right).\left(-b\right).\left(-a\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
Ta có : a + b + c = 0
=> ( a + b + c )3 = 0
=> a3 + b3 + c3 + 3a2b+ 3ab2+ 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc = 0
=> a3 + b3 + c3 + ( 3a2b+ 3ab2 + 3abc ) + ( 3a2c + 3ac2 + 3abc ) + ( 3b2c + 3bc2 + 3abc ) - 3abc = 0
=> a3 + b3 + c3 + 3ab ( a + b + c ) + 3ac ( a + b + c ) + 3bc ( a + b + c ) = 3abc
MÀ a + b + c = 0
=> a3 + b3 + c3 = 3abc (đpcm)
Hok Tốt!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(VT=a^3+b^3+c^3\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b\right)-3abc+3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)+3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)+3abc\)
\(=3abc=VP\) ( do a + b + c = 0 )
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0\)
\(\Leftrightarrow (a+b)^3+c^3-3ab(a+b+c)=0\)
\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0\)
\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)
\(\Rightarrow \left[\begin{matrix} a+b+c=0\\ a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
Với $a^2+b^2+c^2-ab-bc-ac=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow (a-b)^2=(b-c)^2=(c-a)^2=0$
$\Leftrightarrow a=b=c$
Do đó ta có đpcm.
Ta có : a3 + b3 + c3 = 3abc
=> a3 + b3 + c3 -3abc = 0
=> ( a + b )3 - 3ab( a - b ) + c3 -3abc = 0
=> ( a + b )3 + c3- [(3ab( a +b) + 3abc] =0
=> ( a + b + c)[( a + b )2 - (a+b)c + c2 ] - 3ab( a+b +c ) = 0
=> ( a + b + c)( a2 + 2ab + b2 - ac -bc + c2 - 3ab) = 0
=> ( a + b + c )( a2 + b2 + c2 + ab - bc - ac ) = 0
=> a + b + c = 0 ( đpcm ) hoặc có thể là a = b = c ( đpcm ) nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
câu 2:<=>[(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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow a^3+b^3+3a^2b+3ab^2=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà a+b= -c (cmt )
nên \(a^3+b^3+c^3=3abc\left(đpcm\right)\)
\(a+b+c=0\Rightarrow c=-a-b\)
\(\Rightarrow a^3+b^3+c^3=a^3+b^3+\left(-a-b\right)^3=a^3+b^3-a^3-3a^2b-3ab^2-b^3\)
\(=-3a^2b-3ab^2=3ab\left(-a-b\right)=3abc\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề ta có:
a+b+c=0 => c=-(a+b) (1)
Thay (1) vao a^3+b^3+c^3 ta có:
a3+b3+[-(a+b)]3=3ab[-(a+b)]
<=>a3+b3-(a+b)=-3ab(a+b)
<=> a3+ b3- a3 -3a2b- 3ab2- b3= -3a2b- 3ab2
<=> 0= 0
vậy ta có đpcm.
b) có a+b+c = 0
=> a2+b2+c2+2(ab+bc+ac) = 0
mà a2+b2+c2 = 2
=> ab+bc+ac = -1
=> a2b2+b2c2+a2c2 + 2ab2c+2a2bc+2abc2 = 1
=>a2b2+b2c2+a2c2 + 2abc(b+a+c) = 1
=>a2b2+b2c2+a2c2 = 1
Ta bìn phong cái a2+b2+c2 len
đk là
a4+b4+c4 + 2a2b2+2a2c2+2b2c2=4
=> a4+b4+c4 + 2(a2b2+a2c2+b2c2) = 4
mà ở trên là a2b2+b2c2+a2c2 = 1
=> a4+b4+c4 +1 =4
a4+b4+c4 = 3 D
k giùm nha!!!
a3 + b3 + c3 = 3abc
=> a3 + b3 + c3 + 3ab(a + b + c) + 3bc(a + b + c) + 3a(a + b + c) = 3abc
=> a3 + b3 + c3 + (3a2b + 3ab2 + 3abc) + (3b2c + 3bc2 + 3abc) + (3a2c + 3ac2 + 3abc) - 3abc = 0
=> a3 + b3 + c3 + 3a2b + 3ab2 + 3b2c + 3bc2 + 3a2c + 3ac2 + 6abc = 0
=> (a + b + c)3 = 0
=> a + b + c = 0 (đpcm)