a) \(\left(a+b\right)^3+\left(a+b\right)^3\)

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

\(=2\left(a+b\right)\left[\left(a+b\right)^2\left(1-2+1\right)\right]\)

\(=2\left(a+b\right)\)

b)  \(9x^2+6xy+y^2\)

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

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

c)  \(4x^2-25\)

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

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

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

x⁴y⁴ + 64

= x⁴y⁴ + 16x²y² + 64 - 16x²y²

= (x²y² + 8)² - (4xy)²

= (x²y² - 4xy + 8)(x²y² + 4xy + 8)

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

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

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

\(a,\left(a+b\right)^3+\left(a-b\right)^3\)

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

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

\(=2a\left(a^2+3b^2\right)\)

\(b,9x^2+6xy+y^2\)

\(=\left(3x\right)^2+2.3x.y+y^2\)

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

\(c,4x^2-25\)

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

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

(x^2-x+2)^2+(x-2)^2 
= [(x^2-x+2)+(x-2)]^2-2[(x^2-x+2)*(x-2)] (áp dụng (a^2+b^2)=(a+b)^2-2ab 
=(x^2)^2- 2((x^3-3x^2+4x-4) 
=x^4-2x^3+6x^2-8x+8 
 giờ phân tích đa thức 
x^4-2x^3+6x^2+8x-8 
=(x^4-2x^3+2x^2)+(4x^2-8x+8) (cái này làm bài tập nhiêu nhìn ra nhanh) 
=[x^2(x^2-2x+2)]+4(x^2-2x+2) dẹp luôn 
=(x^2-2x+2)(x^2+4) 

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

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

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

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

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

Cảm ơn hai bạn nha nhưng Đường Quỳnh Giang nhầm dấu rồi.😆

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

                    =a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

                        =6a²b+2b³

Áp dụng hđt a³-b³=(a-b)(a²+ab+b²) ấy

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

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

\(=2b\left(3a^2+b^2\right)\)

Ta có : \(\left(a+b\right)^3-\left(a-b\right)^3\) 

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

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=6a^2b+2b^3\)

x^4+6x^2y+9y^2-1
=(x^2+3y)^2-1
=(x^2+3y+1)(x^2+3y-1)

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

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

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

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

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

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

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