Ta có : \(a+b+c+d=0\Leftrightarrow a+d=-\left(b+c\right)\)

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

\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[c^3+b^3+3bc\left(b+c\right)\right]\)

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=3ad\left(b+c\right)-3bc\left(b+c\right)\) (vì a + d = - b - c )

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

\(a)\)\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Leftrightarrow\)\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d+a+b-c-d}{a-b+c-d+a-b-c+d}=\frac{2\left(a+b\right)}{2\left(a-b\right)}=\frac{a+b}{a-b}\) \(\left(1\right)\)

Lại có : 

\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d-a-b+c+d}{a-b+c-d-a+b+c-d}=\frac{2\left(c+d\right)}{2\left(c-d\right)}=\frac{c+d}{c-d}\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)\(\Leftrightarrow\)\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\) \(\left(3\right)\)

Lại có : 

\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\) \(\left(4\right)\)

Từ \(\left(3\right)\) và \(\left(4\right)\) suy ra \(\frac{a}{c}=\frac{b}{d}\) ( đpcm ) 

Chúc bạn học tốt ~ 

\(b)\)\(a^3+b^3+c^3=3abc\)

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

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\) ( vì \(a+b+c=0\) ) 

\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

Vậy ... 

Chúc bạn học tốt ~ 

Phùng Minh Quân: hình như câu b làm nhầm đề rồi đấy.

\(a+b+c=0\)

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

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

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

\(\Rightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

                                   đpcm

Tham khảo nhé~

a) Ta có: (a + b + c + d)(a - b - c +d )=( (a + d) + (b + c) )( (a + d) - (b + c) )

                                                     =(a + d )² - (b +c )²                             (1)

              (a - b + c - d)(a + b - c - d)=(a - d)² - (b - c)²                                  (2)

Từ (1) và (2)  => a² + 2ad + d² - b² - 2bc - c²=a² - 2ad + d² - b² + 2bc - c²

4ad=4bc => ad=bc <=> \(\frac{a}{c}=\frac{b}{d}\)  (đpcm)

\(a.a^3+b^3+c^3=3abc\)

⇔ \(a^3+b^3+c^3-3abc=0\)

⇔ \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

⇔ \(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

⇔ \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Với : a + b + c = 0 thì dễ thấy đẳng thức trên đúng .

Từ đó suy ra : đpcm .

\(b.a+b+c+d=0\)

⇔ \(a+b=-\left(c+d\right)\)

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

⇔ \(a^3+b^3+3a^2b+3ab^2=-\left(c^3+3c^2d+3cd^2+d^3\right)\)

⇔ \(a^3+b^3+c^3+d^3=-3c^2d-3cd^2-3a^2b-3ab^2\)

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

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

⇔ \(a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\) ( đpcm)

\(a^2+b^2=2ab\)

<=>  \(a^2+b^2-2ab=0\)

<=>  \(\left(a-b\right)^2=0\)

<=>   \(a-b=0\)

<=>  \(a=b\)  (đpcm)

\(a^3+b^3+c^3=3abc\)

<=>  \(a^3+b^3+c^3-3abc=0\)

<=>  \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

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

<=>  \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>  \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

Xét:  \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>  \(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

<=>  \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>  \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

<=>  \(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)

<=>  \(a=b=c\)

=>  đpcm

cách khác:

Áp dụng BĐT AM-GM ta đc:

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu "=" xảy ra  <=>  \(a=b=c\)

c)  bạn lm tương tự