sửa đề thành \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

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

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

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

                      \(=\left(a+b\right)\left(ab+bc+a^2+ca\right)\)

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

                       \(=\left(a+b\right)\left[b\left(a+c\right)+c\left(c+a\right)\right]\)

                        \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(A=\left(a+b+c\right).\left(bc+ca+ab\right)-abc\\ =abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\\ =\left(b^2c+bc^2\right)+\left(a^2b+a^2c\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\\ =bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(b+c\right)+ab\left(b+c\right)\\ =\left(b+c\right)\left(bc+a^2+ac+ab\right)\\ =\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=\left(b+c\right)\left(a+c\right)\left(a+b\right)\)

(a + b + c)(bc + ca + ab) − abc

=(a + b)(bc + ca + ab) + c(bc + ca + ab) − abc

=(a + b)(bc + ca + ab)+ abc + c²(a + b) − abc

=(a + b)(bc + ca + ab + c²)

=(a + b)(b + c)(c + a)

\(ab\left(a+b\right)-bc\left(b+c\right)+ca\left(a+c\right)+abc\)

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

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

       Đến đây thì mk chịu

ab(a+b) + bc(b+c) + ca(c+a) = a^2b + ab^2 + b^2c + bc^2 + ca(c+a) + 2abc 
= ab^2 + b^2c + a^2b + bc^2 + 2abc + ca(c+a) 
=b^2(a+c) + b(a^2 + c^2 + 2ac) + ca(c+a) 
=b^2(a+c) + b(a+c)^2 + ca(c+a) 
=(c+a)[b^2 + b(a+c) + ca] 
=(c+a)[b^2 + ab + bc + ca] 
=(c+a)[b(b+a) + c(b+a)] 
=(c+a)(b+c)(b+a) 

 ta có: ab(a + b) + bc(b + c) + ac(a + c) + 3abc 

= ab(a + b) + abc + bc(b + c) + abc + ac(a + c) + abc 

= ab(a + b + c) + bc(a + b + c) + ac(a + b + c) 

= (a + b + c)(ab + bc + ca) 

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

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

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

\(=\left(b+c\right)\)\(\orbr{c\left(b+a\right)-a\left(a+b\right)}\)

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

Tỉ lệ sai 1%

(a+b+c)(ab+bc+ca)−abc

=(a+b)(ab+bc+ac)+c(ab+bc+ca)−abc

=(a+b)(ab+bc+ca)+abc+c2(a+b)−abc

=(a+b)(ab+bc+ca+c2)

=(a+b)(b+c)(c+a)

nguồn: https://h7.net/hoi-dap/toan-8/phan-h-a-b-c-ab-bc-ca-abc-thanh-nhan-tu--faq429360.html

Tham khảo tại đây nhé bạn Nguyễn Hà Anh

https://olm.vn/hoi-dap/detail/10986837094.html

dòng thứ 3 và thứ 4 là c² nhé, ko phải c2

= (abc - ab) + (a - ca) + (b - bc) + (c -1)

= ab.(c -1) - a.(c - 1) - b(c -1) + (c -1)

= (c -1).(ab - a - b  + 1)