\(1-27x^3\)

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

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

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

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

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

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

mày viết lại cái đề bài hộ tao cái

a: \(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+9x\left(x+3\right)=0\)

=>(x+3)(x^2+6x+9)=0

=>x=-3

b: \(\Leftrightarrow x^2-2x-x^3+6x^2-4=0\)

=>-x^3+6x^2-2x-4=0

hay \(x\in\left\{5.5;1.14;-0.64\right\}\)

c: =>(3x-1)^3=8

=>3x-1=2

=>3x=3

=>x=1

\(=10^3=1000\)

=10³=1000

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

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

\(x^3-9x^2+27x-27\)

=\(x^3-3.x^2.3+3.x.3^3-3^3\)

\(\left(x-3\right)^3\)

Thay x=1, ta có:

(1-3)³=-2³=-8

\(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow\left(x+3\right)^3=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

KL:..............

x³ + 9x² + 27x + 27 = 0

<=> x² + 3.3.x² + 3.3².x + 3³ = 0

<=> (x + 3)³ = 0

<=> x + 3 = 0

<=> x = -3

x³ + 9x² + 27x +27 = 0

=> x³ + 3.x².3 + 3.x.3² + 3³ = 0

=>( x + 3)³ = 0

=> x + 3 = 0

=> x = -3

Vậy x = -3

\(x^3+9x^2+27x+27=0\)

\(x^3+3\times x^2\times3+3\times x\times3^2+3^3=0\)

\(\left(x+3\right)^3=0\)

\(x+3=0\)

\(x=-3\)