Ta có

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=0\Rightarrow ab+bc+ac=0.\)

\(A=\frac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^2}\)

Ta có

\(\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3-3\left(abc\right)^2=\)

\(=\left(ab+bc+ac\right)\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2-abbc-bcac-abac\right]=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3\left(abc\right)^2\)

\(\Rightarrow A=\frac{3\left(abc\right)^2}{\left(abc\right)^2}=3\)

`(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`

`VT>=0`

Dấu "=" xảy ra khi `a=b=c`

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

`<=>a^3+b^3+c^3-3abc=0`

`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`

`<=>(a+b)^3+c^3-3ab(a+b+c)=0`

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

`**a+b+c=0`

`**a^2+b^2+c^2=ab+bc+ca`

`<=>a=b=c`

Bài 2: 

a+b+c+d=0

nên b+c=-(a+d)

\(a^3+b^3+c^3+d^3\)

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

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

\(=3ad\left(b+c\right)-3bc\left(b+c\right)\)

\(=\left(b+c\right)\left(3ad-3bc\right)\)

\(=3\left(b+c\right)\left(ad-bc\right)\)

a) a² + b² + c2 = ab + ac + bc

=> 2a² + 2b² + 2c² = 2ab + 2ac + 2bc

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 0

=> (a - b)² + (a - c)² + (b - c)² = 0 

Do 3 hạng tử trên đều có giá trị lớn hơn hoặc bằng 0 nên a - b = a - c = b - c = 0

=> a = b = c 

b) a³ + b³ + c³ = 3abc

=> a³ + b³ + c³ - 3abc = 0

=> a³ + 3a²b + 3ab² + b³ + c³ - 3abc - 3a²b - 3ab² = 0

=> (a + b)³ + c³ - 3ab(a + b + c) = 0

=> (a + b + c)(a² + 2ab + b² - bc - ac + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - bc - ac) = 0 

=> a + b + c = 0

hoặc a² + b² + c² = ab + bc + ac =>  a = b = c

a)\(a^2+b^2+c^2=ab+bc+ca\)\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

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

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

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

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

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

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

\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

Bài 2:

Ta có: \(a+b+c=0\Rightarrow a+b=-c\)

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

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

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

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

(Còn nhiều cách nữa ,mình làm 1 cách nhé)

\(Xét:a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3=-3ab\left(a+b\right)=3-ab\left(-c\right)=3abc\Rightarrow a^3+b^3+c^3=3abc\Rightarrow E=0\)

ngoại trừ với: a=b=c thì phân số vô nghĩa không cần cho điều kiện a+b+c=0 thì ta cũng phân tích:

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