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

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

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

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

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

=(x-y-2y)[(x-y)^2+2y(x-y)+4y^2]

=(x-3y)(x^2-2xy+y^2+2xy-2y^2+4y^2)

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

\(\left(x-y\right)^3-8y^3\)

\(=\left(x-y\right)^3-\left(2y\right)^3\)

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

\(=\left(x-y-2y\right)\left(x^2-2xy+y^2+2xy-2y^2+4y^2\right)\)

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

\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)

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

\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)

\(=\left(x-y\right)\left(3x^2-3xy+6\right)\)

(x - y)(3x² - 3xy + 6)

a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)

\(=\left(2x-y\right)^3\)

\(x^3-6x^2y+12xy^2-8y^3\)

\(=\left(x-2y\right)^3\)

 =(y+x-6)(y+x-2)

Đặt \(x+y=u\)

Biểu thức trở thành \(u^2-8u+12\)

\(=u^2-2u-6u+12\)

\(=u\left(u-2\right)-6\left(u-2\right)\)

\(=\left(u-6\right)\left(u-2\right)\)

Thay ngược trở lại, ta được:

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

\(x^2+2xy+7x+7y+y^{2+10}\)

\(\text{phân tích đa thức thành nhân tử}\)

\(y^{12}+2xy+7y+x^2+7x\)

tách \(^{x^2}\)ra rồi làm thừa số chung, toán SGK đem ra hỏi làm j

\(x^2-y^2+8x+6y+7\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)

\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)

\(=\left(x+y+1\right)\left(x-y+7\right)\)

Ta có x² - y² + 8x + 6y + 7

= x² + 8x + 16 - y² + 6y - 9

= \(x^2+4x+4x+16-y^2+3y+3y-9\)

= x(x + 4) + 4(x + 4) - y(y - 3) + 3(y - 3)

= (x + 4)² - (y - 3)²

= (x + 4 + y - 3)(x + 4 - y + 3) 

= (x + y + 1)(x - y + 7)

x^4 + y^4=(x^2)^2+(y^2)^2
=(x^2+y^2)^2-2x^2y^2
=(x^2+y^2)^2-(√2xy)^2
=(x^2+y^2-√2 xy)(x^2+y^2+√2 xy)