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

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

(b-c)^2+b(a-c)^2+c(a-b)^2- a^3 -b^3 -c^3 +4abc 
=a[(b-c)^2-a^2)]+ b[(a-c)^2-b^2)]+c[(a-b)^2-c^2)]+4abc 
=a[(b-c)^2-a^2)]+ b[(a+c)^2-b^2)]+c[(a-b)^2-c^2)] 
=a(b-c-a)(b-c+a)+b(a+c-b)(a+b+c)+c(a+c... 
=[-a(b-c+a)+b(a+b+c)+c(a-b-c)](a+c-b) 
Bạn cứ tiếp tục phân tích cái vế trong ngoặc vuông đuọc (a+b-c)(b+c-a) là đc.

Đáp số : (a+c-b)(a+b-c)(b+c-a) 

:)) Thớt không search google nên bạn í search hộ thôi =.=

Đinh Tuấn Việt- sai dấu ở hàng thứ 3, còn 4abc nằm ở đâu?

\(a,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-40\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)=\left(x^2+6x\right)\left(x^2+6x+13\right)\\ b,=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\\ =a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)-b^2c^2\left(c-a\right)+c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(a^2b^2-b^2c^2\right)-\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)-c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(a-c\right)\left[b^2\left(a+c\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(a-c\right)\left(a^2b+b^2c-b^2c+a^2c\right)\\ =a^2\left(a-b\right)\left(a-c\right)\left(b+c\right)\)

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

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

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

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

b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c

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

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

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

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

=(c-a)[b(b+c)+a(b+c)]

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

$A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1$

A = 8abc + 4ab + 4bc + 4ca + 2a + 2b + 2c + 1

$A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)$

$A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)$

$A=\left(2c+1\right)\left(4ab+2a+2b+1\right)$

$A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]$

$A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)$

=a^2b-a^2c+b^2a-b^2c+c^2a-c^2b

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

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

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

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

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

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

= (ab - b - a + 1)(c - 1) 

= (a - 1).(b - 1).(c - 1)   

\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)