ta có : 

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

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

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

b. tương tự ta có :

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

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

c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

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

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

mình nè

bạn gửi lời kết bạn nhé mình hết ượt rui ok 

Cậu gửi lời mời kb nha.

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!!!

a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\)    *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\)  cho nhanh*

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

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

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

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

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

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

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

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

a,(x+1)²

b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}

c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}

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

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

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

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

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

 Thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

Biến đổi vế trái thành: 

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)

boc vai

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)

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