- Phân tích ra nhân tử :

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

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

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

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

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

Từ đây ta có \(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-bc-ac}\)

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

\(\frac{a^3+b^3-c^3+3abc}{\left(a-b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}=\frac{\left(a+b\right)^3-c^3-3ab\left(a+b\right)+3abc}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)

                                                                     \(=\frac{\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)

                                                                     \(=\frac{\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)}{2\left(a^2+b^2+c^2-ab+bc+ac\right)}\)

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

\(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

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

\(=\frac{\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc-c^2\right)-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc-c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)

a^3+b^3+c^3-3abc

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc
<=>[(a+b)^3 +c^3] -3ab.(a+b+c)

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)

<=>(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)

thay vào và rút gọn ta được:\(a+b+c\)

a^3 +c^3 = (a+c). (a^2 -a.c+c^2)

 = (a+c)^3 -3 ac.(a+c)

 => a^3+c^3-3abc+b^3 =(a+c)^3-3ac (a+c)-3abc +b^3

=(a+c)^3+b^3 -3ac (b+(a+c))

=(a+c+b). ((a+c)^2-(a+c).b+b^2) -3ac (a+c+b)

 =(a+c+b)^3-3(a+c)b. (a+c+b)-3ac (a+c+b)

 =(a+c+b)((a+c+b)^2  -3ab-3bc-3ac) (1)

 (a-b)^2 + (b-c)^2 +(a-c)^2 

 = 2a^2 +2b^2+2c^2 -2ab-2bc-2ac

 =2 (a^2+b^2+c^2-ac-ab-bc)

 =2((a+b)^2-3ab +c^2 -ac-bc)

 =2 ((a+b+c)^2-2(ac+bc)-3ab-ac-bc)

 =2 (( a+c+b)^2 -3ab-3bc -3ac) (2)

Từ (1),(2) =>(a^3+b^3+c^3-3abc)/((a-b)^2

+(b-c)^2+(c-a)^2)

=(a+b+c)/2 

M=a^3+b^3+c^3-3abc/(a-b)^3+(b-c)^3+(c-a)^3

nè ban

a, Gợi ý nà :3

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

a^2 - b^2 + c^2 + 2ac = (a + c)^2 - b^2 = (a + b + c)(a - b + c)

b. Gợi ý tiếp luôn nà :3

a^3 + b^3 + c^3 - 3abc

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

= (a+b)^3 + c^3 - 3ab(a+b+c) 

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

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

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

Rồi cứ thế rút gọn...

Học tốt nha bạn :3

\(\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\frac{a+b-c}{a-b+c}\)

\(\text{nhận xét: ta có hằng đẳng thức:}\)

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

đó đến đây bạn làm tiếp

b/\((\sum a^3)-3abc=(\sum a).(\sum a^2-\sum ab)\)\(\Rightarrow\)\(\frac{(\sum a^3)-3abc}{(\sum a^2-\sum ab)}=\frac {(\sum a).(\sum a^2-\sum ab)}{(\sum a^2-\sum ab)}=a+b+c\)

a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)