a³ + b³ + c³ - 3abc = (a + b)³ + c³ - 3abc - 3ab(a + b)

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

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

\(a^3+b^3+c^3-3abc\)

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

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

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

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

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

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

\(=xy\left(x^2-2xy+5\right)\)

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

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

`a, 8x^3 - 1 = (2x-1)(4x^2 + 2x - 1)`

`b, x^3 + 27y^3 = (x+3y)(x^3 - 3xy + 9y^2)`

`c, x^3 - y^6 = (x-y^2)(x+xy^2 + y^4)`

a) \(x^2-9+2\left(x+3\right)=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x-3+2\right)=\left(x+3\right)\left(x-1\right)\)

b) \(x^2-10x+25-3\left(x-5\right)=\left(x-5\right)^2-3\left(x-5\right)=\left(x-5\right)\left(x-5-3\right)=\left(x-5\right)\left(x-8\right)\)

c) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)

Ta có

a³+b³+c³-3abc

=(a+b)³-3ab(a+b)+c³-3abc

=[(a+b)³+c³]-3ab(a+b+c)

=(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)

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

=(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)

=(a+b+c)(a²+b²+c²-ab-ac-bc)

a³+b³+c³-3abc

=(a+b)³-3ab(a+b)+c³-3abc

=[(a+b)³+c³]-3ab(a+b+c)

=(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)

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

=(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)

=(a+b+c)(a²+b²+c²-ab-ac-bc)

a³ + b³ + c³ - 3abc

= [( a³ + b³ ) + c³ ] - 3abc

= [( a + b )³ + c³ + 3ab( a + b )] - 3abc

= [( a + b )³ + c³ ] + 3a²b + 3ab² - 3abc

= ( a + b + c ) [( a + b )² - c( a + b ) + c² ] - 3ab( a + b + c )

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

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

= 1/2( a + b + c ) ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ac )

= 1/2( a + b + c ) [( a² - 2ab + b² ) + ( a² - 2ac  + c² ) + ( b² - 2bc + c² )]

= 1/2 ( a + b + c ) [( a - b )² + ( a - c )² + ( b - c )²]

Hok Tốt!!!

M = (a + b + c)³ - a³ - b³ - c³

= (a + b)³ + c³ + 3(a + b)²c + 3(a + b)c² - a³ - b³ - c³

= a³ + b³ + c³ + 3a²b + 3ab² + 3(a + b)c(a + b + c) - a³ - b³ - c³

= 3ab (a + b) + 3c(a + b)(a + b + c)

= 3(a + b)[ab + c(a + b + c)]

= 3(a + b)(ab + bc + ac + c²)

= 3(a + b)[b(a + c) + c(a + c)]

= 3(a + b)(b + c)(c + a)

N = a³ + b³ + c³ - 3abc

= (a + b)³ + c³ - 3ab(a + b) - 3abc

= (a + b + c)³ - 3(a + b)c(a + b + c) - 3ab(a + b + c)

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

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

= (a + b + c)(a² + b² + c² - ab - bc - ca)