\(=4x^2y^2-12x^3y^4+9x^2y^6\)

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

=2x-1

a, \(\dfrac{27}{8x^3-1}:\dfrac{3}{2x-1}\)

\(=\dfrac{27}{\left(2x-1\right)\left(4x^2+2x+1\right)}.\dfrac{2x-1}{3}\)

\(=\dfrac{9}{4x^2+2x+1}\)

b, \(\dfrac{8x^3+36x^2+54x+27}{2x+3}=\dfrac{\left(2x+3\right)^3}{2x+3}=\left(2x+3\right)^2\)

\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)

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

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

\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)

a: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

hay x=1/2

b: \(\Leftrightarrow\left(3x-1\right)^2=0\)

hay x=1/3

c: \(\Leftrightarrow\left(x+4\right)^2=0\)

hay x=-4

a) ⇒ \(\left(x-\dfrac{1}{2}\right)^2\)= 0

⇒ \(x-\dfrac{1}{2}=0\)

⇒ \(x=\dfrac{1}{2}\)

b) ⇒ \(\left(3x-1\right)^2=0\)

⇒ \(3x-1=0\)

⇒ \(3x=1\)

⇒ \(x=\dfrac{1}{3}\)

c) ⇒ \(\left(x+4\right)^2=0\)

⇒ \(x+4=0\)

⇒ \(x=-4\)

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

2, đề sai 

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

4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)

tương tự ... 

8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

1) \(x^3-1=x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)\)

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

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

Bài 1 : \(x^3-1=\left(x-1\right)\cdot\left(x^2+x+1\right)\)

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

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