\(1-27x^3\)

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

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

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

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

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

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

\(---\)

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

P = ( m² - 2m + 4 )( m + 2 ) - m³( m + 3 )( m - 3 ) - m² - 18

= m³ + 8 - m³( m² - 9 ) - m² - 18

= m³ + 8 - m⁵ + 9m³ - m² - 18

= -m⁵ + 10m² - m² - 10

N = ( x + y )³ - 9( x + y )² + 27( x + y ) - 27

= ( x + y )³ - 3.( x + y )².3 + 3.( x + y ).3² - 3³

= ( x + y - 3 )³

Phụ thuộc vào biến hết mà ;-;

\(P=\left(m^2-2m+4\right)\left(m+2\right)-m^3+\left(m+3\right)\left(m-3\right)-m^2-18\)

\(=m^3+8-m^3+\left(m^2-9\right)-m^2-18\)

\(=m^3+8-m^3+m^2-9-m^2-18\)

\(=-19\)

Vậy biểu thức trên kh thụ vào biến m

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

\(=\left(x+y\right)^3-3\left(x+y\right)^2.3+3\left(x+y\right)3^2-3^3\)

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

\(D=\left(2+6\right)^3-9\left(2+6\right)^2+27\left(2+6\right)-27\)

    \(=8^3-3\cdot3\cdot8^2+3\cdot3^2\cdot8-3^3\)

     \(\left(8-3\right)^3=5^3=125\)

a) 

\(3x^2-x^3-9x+3x^2+27-9x=27-x^3\) 

\(-x^3+6x^2-18x+27=27-x^3\) 

\(6x^2-18x=0\) 

\(6x\left(x-3\right)=0\) 

\(\orbr{\begin{cases}6x=0\\x-3=0\end{cases}}\) 

\(\orbr{\begin{cases}x=0\\x=3\end{cases}}\) 

b) 

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

\(x^4-y^4=x^4-y^4\) 

\(0=0\left(llđ\forall x\right)\) 

a) ( x² - 3x + 9 )( 3 - x ) = 27 - x³

<=> -x³ + 6x² - 18x + 27 = 27 - x³

<=> -x³ + 6x² - 18x + x³ = 27 - 27

<=> 6x² - 18x = 0

<=> 6x( x - 3 ) = 0

<=> \(\orbr{\begin{cases}6x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

b) Ta có VP = ( x² )² - ( y² )²

                    = ( x² - y² )( x² + y² )

                    = ( x - y )( x + y )( x² + y² )

                    = ( x - y )[ ( x + y )( x² + y² ) ]

                    = ( x - y )( x³ + xy² + x²y + y³ ) = VT

Vậy phương trình nghiệm đúng với mọi x, y ∈ R

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

\(27-x^3=27-x^3\)

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

\(VP=x^4-y^4\)

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

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

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

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

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