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

(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)

\(A=\left(a+b+c\right)\left(bc+ac+ab\right)-abc\)

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

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

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

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

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

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

= (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)

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

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

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

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

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

=(c-1)[b.(a-1)-(a-1)]

=(c-1)(a-1)(b-1)

ai biết trả lời nhanh hộ mình nha! Mình k đúng cho!

Co P=ab(a-b) + bc((b-a)+(a-c)) +ac(c-a) 
=ab(a-b) -bc(a-b) -bc(c-a) +ac(c-a) 
=(a-b)(ab-bc) +(c-a)(ac-bc) 
=(a-b) b (a-c) + (c-a) c (a-b) 
=(a-b)(a-c)(b-c) 

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