a: =xy(x^2-4xy^2+4y^4)

=xy(x-2y^2)^2

b:=(x^3-y)^2

c: =(a^2-b^2)(a^2+b^2)

=(a^2+b^2)(a-b)(a+b)

d: 64x^6-27y^6

=(4x^2-3y^2)(16x^4+12x^2y^2+9y^4)

e: =(2x)^3+(3y)^3

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

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

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

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

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

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

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

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: Ta có: \(x^{10}-4x^8+4x^6\)

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

\(=x^6\left(x-2\right)^2\left(x+2\right)^2\)

2: Ta có: \(m^3+27\)

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

3: Ta có: \(x^3+8\)

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

4: Ta có: \(\frac{1}{27}+a^3\)

\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)

5: Ta có: \(8x^3+27y^3\)

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

6: Ta có: \(\frac{1}{8}x^3+8y^3\)

\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

7: Ta có: \(8x^6-27y^3\)

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

8: Ta có: \(\frac{1}{8}x^3-8\)

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

9: Ta có: \(\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

10: Ta có: \(\left(a+b\right)^3-c^3\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)

\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)

11: Ta có: \(x^3-\left(y-1\right)^3\)

\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)

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

12: Ta có: \(x^6+1\)

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

1) \(x^{10}-4x^8+4x^6\)

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

2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)

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

4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)

5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

8) \(\frac{1}{8}x^3-8=\left(\frac{1}{2}x\right)^3-2^3=\left(\frac{1}{2}x-2\right)\left[\left(\frac{1}{2}x\right)^2+\frac{1}{2}x.2+2^2\right]=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)

10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)

P/s: Đăng ít thôi chớ bạn!

b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

c: \(4x^2+12x+9=\left(2x+3\right)^2\)

d: \(25x^2-20xy+4y^2=\left(5x-2y\right)^2\)

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

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)

a) \(8x^3-64\)

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

\(=\left(2x-4\right)\left[\left(2x\right)^2+2x.4+4^2\right]\)

\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)

d) \(125x^3+27y^3\)

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

\(=\left(5x+3y\right)\left[\left(5x\right)^2-5x.3y+\left(3y\right)^2\right]\)

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

Câu b có sai đề không ạ ?

`a)8x^3+27y^3`

`=(2x)^3+(3y)^3`

`=(2x+3y)(4x^2-6xy+9y^2)`

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

\(\Leftrightarrow x:3+367\cdot\left(-2\right)=-60\)

=>x:3=674

hay x=2022

x= 2022

\(8x^3-64\)

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

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