a: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

b: \(27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)

c: \(y^6+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)

d: \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

giúp mình câu e,f với

a) \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

b) \(125x^6-27x^9=\left(5x^2-3x^3\right)\left(25x^4+15x^5+9x^6\right)\)

c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

a: \(8x^3-1=\left(2x-1\right)\left(4x^2+2x+1\right)\)

b: \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

c: \(x^3+125=\left(x+5\right)\left(x^2-5x+25\right)\)

d: \(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

a) 8x³ - 1

= (2x)³ - 1³

= (2x - 1)(4x² + 2x + 1)

b) x³ + 8y³

= x³ + (2y)³

= (x + 2y)(x² + 2xy + 4y²)

c) x³ + 125

= x³ + 5³

= (x + 5)(x² - 5x + 25)

d) x³ - 27y³

= x³ - (3y)³

= (x - 3y)(x² + 3xy + 9y²)

 Chúc bạn học tốt

a. 8x³ - 1

⇔ (2x)³ - 1³

⇔ (2x - 1)³.

b. x³ + 8y³

⇔ x³ + (2y)³

⇔ (x + 2y)³

c. x³ + 125

⇔ x³ + 5³

⇔ (x + 5)³

d. x³ - 27y³

⇔ x³ - (3y)³

⇔ (x - 3y)³

`a, (2x-3)^3 = 8x^3 - 36x^2 + 54x - 27`

`b, (a+3b)^3 = a^3 + 9a^2b + 27ab^2 + 27b^3`

`c, (xy-1)^3 = x^3y^3 - 3x^2y^2 + 3xy -1`

\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)

a: \(3abc^3-6a^2b^3c+12a^3bc\)

\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)

\(=3abc\left(c^2-2ab^2+4a^2\right)\)

b: \(27-8y^3\)

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

\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)

c: Sửa đề: \(4x^2+4x-y^2+1\)

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

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

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

d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)

\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)

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

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

\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

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

b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)

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

b/ \(64x^3+0,001=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\)

c/ \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d/ \(\dfrac{x^3}{125}-\dfrac{y^3}{27}=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

`a, (a-1)(a+1)(a^2+1)`

`= (a^2-1)(a^2+1)`

`= a^4-1`

`b, (xy+1)^2 - (xy-1)^2`

`= x^2y^2 + 2xy + 1 - x^2y^2 + 2xy - 1`

`= 4xy`

a) \(\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\)

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

\(=a^4-1\)

b) \(\left(xy+1\right)^2-\left(xy-1\right)^2\)

\(=\left[\left(xy+1\right)-\left(xy-1\right)\right]\left[\left(xy+1\right)+\left(xy-1\right)\right]\)

\(=\left(xy+1-xy+1\right)\left(xy+1+xy-1\right)\)

\(=4xy\)

a: \(1-\dfrac{x^3}{8}=\left(1-\dfrac{1}{2}x\right)\left(1+\dfrac{1}{2}x+\dfrac{1}{4}x^2\right)\)

b: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)

c: \(64x^3-27y^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)