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c, Ta có : a+b+c=0 ⇒ c=-(a+b)
⇒ a3+b3+c3= a3+b3-(a+b)3= x3+y3-(x3+3x2y+3xy2+y3)= x3+y3-x3-3x2y-3xy2-y3= -3x2y-3xy2= -3xy(x+y)= 3xyz(đpcm)
Câu a : Ta có :
\(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z-xyz+y^2z\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\)
Câu b : Khai triển VT ta có :
\(VT=\left(a+b+c\right)^3-a^3-b^3-c^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có :
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Luôn đúng vì \(a+b+c=0\)
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a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)
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Giải:
\(a+b+c+d=0\)
\(\Leftrightarrow a+c=-b-d\)
\(\Leftrightarrow a+c=-\left(b+d\right)\)
Ta có:
\(\left(a+c\right)^3=-\left(b+d\right)^3\)
\(\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-\left(b^3+3b^2d+3bd^2+d^3\right)\)
\(\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\)
\(\Leftrightarrow a^3+3ac\left(a+c\right)+c^3=-b^3-3cd\left(b+d\right)-d^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)-3ac\left(a+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)+3ac\left(b+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
Vậy ...
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Lời giải:
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow (a+b+c)^3-3(a+b)(b+c)(c+a)=3abc\)
\(\Leftrightarrow (a+b+c)^3-3[(a+b+c)(ab+bc+ac)-abc]=3abc\)
\(\Leftrightarrow (a+b+c)^3-3(a+b+c)(ab+bc+ac)=0\)
\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)
Vì \(a,b,c>0\Rightarrow a+b+c>0\)
Do đó \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow 2(a^2+b^2+c^2-ab-bc-ac)=0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)
Ta thấy \((a-b^2;(b-c)^2;(c-a)^2\geq 0\), do đó điều trên xảy ra khi mà:
\(\left\{\begin{matrix}
(a-b)^2=0\\
(b-c)^2=0\\
(c-a)^2=0\end{matrix}\right.\Leftrightarrow a=b=c\)
Ta có đpcm.
\(\text{Ta có }:a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b+c\right)^2=0\)
\(Do\left(a-b\right)^2\ge0\forall x\\ \left(a-c\right)^2\ge0\forall x\\ \left(b-c\right)^2\ge0\forall x\\ \Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b+c\right)^2\ge0\forall x\)
\(\text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Leftrightarrow a=b=c\)
Vậy \(a=b=c\text{ }khi\text{ }a^3+b^3+c^3=3abc\)
Cho các số a, b, c thỏa mãn a^3+ b^3+ c^3= 3abc với a, b, c khác 0. Chứng minh a+ b+c = 0 hoặc a=b=c
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a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 - 3abc = 0
⇒ ( 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 + b2 + c2 - ab - bc - ac ) = 0
⇒ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)
+) a2 + b2 + c2 - ab - bc - ac = 0
⇒ 2( a2 + b2 + c2 - ab - bc - ac ) = 2.0
⇒ 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
⇒ ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( a2 - 2ac + c2 ) = 0
⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 = 0
VT ≥ 0 ∀ a,b,c . Dấu "=" xảy ra khi a = b = c
⇒ a + b + c = 0 hoặc a = b = c ( đpcm )
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Ta có: a + b + c = 0
⇒ a + b = -c ⇒ (a + b)3 = (-c)3
⇒ a3 + b3 + 3ab(a + b) = -c3 ⇒ a3 + b3 + 3ab(-c) + c3 = 0
⇒ a3 + b3 + c3 = 3abc
a+b+c=0
\(\Rightarrow\)(a+b+c)\(^3\)=0
\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+3a\(^2\)b+3ab\(^2\)+3b\(^2\)c+3bc\(^2\)+3a\(^2\)c+3ac\(^2\)+6abc=0
\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+(3a\(^2\)b+3ab\(^2\)+3abc)+(3b\(^2\)c+3bc\(^2\)+3abc)+(3a\(^2\)c+3ac\(^2\)+3abc)-3abc=0
=>a\(^3\)+b\(^3\)+c\(^3\)+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a\(^3\)+b\(^3\)+c\(^3\)=3abc(ĐPCM)
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\(\frac{a^3+b^3+c^3-3abc}{a+b+c}=\frac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a+b+c}=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a+b+c}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a+b+c}=a^2+b^2+c^2-ab-bc-ca\)
\(=\frac{1}{2}\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)\)
\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\) (đpcm)