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 \(a+b+c=0\)

 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)

hey you, còn câu b,c?

ở đây có ai thích sơn tùng không ?

 Ta có \(\dfrac{a^3+b^3}{2ab}\ge\dfrac{ab\left(a+b\right)}{2ab}=\dfrac{a+b}{2}\) 

(áp dụng BĐT quen thuộc \(a^3+b^3\ge ab\left(a+b\right)\))

 Lập 2 BĐT tương tự rồi cộng theo vế:

 \(VT\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\)

 Dấu "=" xảy ra khi \(a=b=c\)

 Ta có đpcm.

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 ...

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 \(a=b=c\text{ }khi\text{ }a^3+b^3+c^3=3abc\)