1, x³-9x²y+27xy²-27y³=(x-3y)³

2, 27x³-9x²y+xy²-\(\dfrac{1}{27}\)y³=(3x-\(\dfrac{1}{3}\)y)³

3)x⁶-3x⁴y+3xy²-y³=(x²-y)³

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

2) \(27x^3-9x^2y+xy^2-\dfrac{1}{27}y^3=\left(3x-\dfrac{1}{3}y\right)^3\)

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

A) \(x^3+27\)

\(=x^3+3^3\)

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

B) \(x^3-\dfrac{1}{8}\)

\(=x^3-\left(\dfrac{1}{2}\right)^3\)

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

C) \(8x^3+y^3\)

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

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

D) \(8x^3-27y^3\)

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

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

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

b)\(\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)

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

d)\(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)

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

c: \(\dfrac{1}{8}x^3-\dfrac{9}{4}x^2y+\dfrac{27}{2}xy^2-27y^3=\left(\dfrac{1}{2}x-3y\right)^3\)

b: \(-x^3+12x^2-48x+64=\left(-x+4\right)^3\)

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

b: =(3x+1)(9x^2-3x+1)

c: =(x+3)(x^2-3x+9)

d: =(4x-3y)(16x^2+24xy+9y^2)

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

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

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

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

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

\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)

a) Ta có: \(x-2y=-4\Rightarrow\left(x-2y\right)^2=16\)

\(\Rightarrow x^2-4xy+4y^2=16\Rightarrow x^2+4y^2=16+4xy=16+4.6=40\)

\(x^3-8y^3=\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(-4\right)\left(40+2.6\right)=-208\)

b) Ta có: \(x+3y=10\Rightarrow x^2+6xy+9y^2=100\Rightarrow x^2+9y^2=100-6xy=100-6.3=82\)

 \(x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=10\left(82-3.3\right)=730\)

\(\dfrac{1}{3}x+y+1=0\)

=>\(\dfrac{1}{3}x+y=-1\)

\(M=x^3+9x^2y+27xy^2+27y^3+27\)

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

\(=\left(x+3y\right)^3+27\)

\(=\left[3\left(x+\dfrac{1}{3}y\right)\right]^3+27\)

\(=27\left(x+\dfrac{1}{3}y\right)^3+27\)

\(=27\left(-1\right)^3+27=0\)

Bạn gõ latex đc ko, cái này này chứ đề khó hiểu quá