= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

x^3-3x^2+3x-1 = (x^3+1)-(3x^2-3x)

(Mình sẽ có hằng đẳng thức x^3+1 cũng giống như x^3+1^3 vì 1^3=1 nhé ) 

= ( x^3+1^3)- (3x^2-3x )

=(x-1)* (x^2+ x*1 + 1^2) -( 3x^2-3x)( Chuyển sang hằng đăng thức ) 

=(x-1 ) *(x^2+ x + 1 ) - 3x(x+1)

=(x-1)*(x^2+x+1-3x ) 

 CÓ MỘT BƯỚC LÀ VÌ DẤU TRỪ Ở TRƯỚC NÊN ĐỔI X+1 THÀNH X-1 NHÉ 

Nếu đúng k dùm minha j , cảm ơn

\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)

\(x^3-3x^2y+x+3xy^2-y-y^3\\=(x^3-3x^2y+3xy^2-y^3)+(x-y)\\=(x-y)^3+(x-y)\\=(x-y)[(x-y)^2+1]\\=(x-y)(x^2-2xy+y^2+1)\)

\(x^3+3x^2+3x+1-27x^3=-26x^3+3x^2+3x+1\)

\(=-26x^3+13x^2-10x^2+5x-2x+1\)

\(=\left(2x-1\right).\left(-13x^2-5x-1\right)\)

3x(x-1)²-(1-x)³=3x(x-1)²-(1-x)².(1-x)=3x(x-1)²+(x-1)²​.(1-x)=(x-1)²(3x-1-x)=(2x-1)(x-1)²

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

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

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

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2x\cdot2y\)

\(=4xy\)

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

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

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

\(=2x\cdot\left(x^2+3y^2\right)\)

______

\(x^3+y^3+z^3-3xyz\)

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