** Sửa: $125x^3-5x-3y+27y^3$
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Lời giải:

$125x^3-5x-3y+27y^3=(125x^3+27y^3)-(5x+3y)$

$=[(5x)^3+(3y)^3]-(5x+3y)$

$=(5x+3y)(25x^2-15xy+9y^2)-(5x+3y)$

$=(5x+3y)(25x^2-15xy+9y^2-1)$

\(8x^3-64\)

\(=8x^3-8^2\)

\(=8\left(x^3-2\right)\)

b) \(125x^3+27y^3=\left(5x\right)^3+\left(3y\right)^3\)

\(=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

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

Ta có

a, a³+b³=(a+b)(a²-ab+b²)

b,x⁶+1=(x²+1)(x²-x+1)

c,125x³-1/8=(5x)³-1/2³=(5x-1/2)(25x²+2,5x+1/4)

d,0,001x³-27y³=(0,1)³-(3y)³=(0,1-3y)(0,01+0,3y+y²)

k cho mính nha

D) 64x^3-1/8y^3

= (4x)^3 + (1/2y)^3

= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]

E) 125x^6-27y^9

( câu này mik chưa rõ nên vx chưa tek giải cho bn )

HOk tốt nhé

a.x^3 + 8

= x^3 + 2^3

= ( x+2) ( x^2 - x.2+2^2)

b/27-8y^3

3^3 - (2y)^3

= ( 3 - (2y))(3^2 + 3.2y + (2y)^2)

c/ y^6 + 1

= (y^3)^3 + 1 ^3

= (y^3 + 1)((y^3)^2 - y^3.1 + 1^2

a) 8x³ - 64 = (2x)³ - 4³

= (2x - 4)\([\)(2x)² + 2x.4 + 4²\(]\)

= (2x - 4)(4x² + 8x + 16)

b) 1 + 8x⁶y³

= 1³ + (2x²y)³

= (1 + 2x²y)[(2x²y)² - 2x²y.1 + 1²]

= (1 + 2x²y)(4x⁴y² - 2x²y + 1)

c) 27x³ + \(\frac{y^3}{8}\)

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

= \(\left(3x+\frac{y}{2}\right)\left[\left(3x\right)^2-3x.\frac{y}{2}+\left(\frac{y}{2}\right)^2\right]\)

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

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)[(5x)² - 5x.3y + (3y)²]

= (5x + 3y)(25x² - 15xy + 9y²)

Câu b có sai đề không ạ ?