\(x^2+5x+6=\left(x^2+2x\right)+\left(3x+6\right)=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)

\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)

\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)

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

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

\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)

Đặt \(x^2+3x+1=a\)ta có :

\(a\left(a+1\right)-6\)

\(=a^2+a-6\)

\(=a^2+6a-a-6\)

\(=\left(a^2+6a\right)-\left(a+6\right)\)

\(=a\left(a+6\right)-\left(a+6\right)\)

\(=\left(a+6\right)\left(a-1\right)\)

Thay \(a=x^2+3x+1\)vào A ta có :

\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)

\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)

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

Còn 1 câu bên dưới nữa b

= (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]

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

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

\(=2\cdot6x\)

\(=12x\)

_________

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

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

\(=2x\cdot2y\)

\(=4xy\)

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

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

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

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

______

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

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

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

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

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

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-1)²-(3x-4)²=(x²+2x+1)-(9x²-24x+16)=-8x²+26x-15=\(-8\left(x-\dfrac{5}{2}\right)\left(x-\dfrac{3}{4}\right)\)