a) Áp dụng HĐT 5 thu được ( 2 a   -   3 b ) 3 .

b) Ta có  8 x 3   +   12 x 2 y   +   6 xy 2   +   y 3  = ( 2 x   +   y ) 3 .

Áp dụng HĐT 7 với A = 2x + y; B = z

( 2 x   +   y ) 3 - z 3 = (2x + y - z)(4 x 2   +   y 2   +   z 2  + 4xy + 2xz + zy).

d) (8a3 – 27b3) – 2a(4a2 – 9b2)

= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)

= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)

b) 8a3 + 4a2b - 2ab2 – b3 = (8a3 – b3 ) + (4a2b - 2ab2 )

= (2a – b)(4a2 + 2ab + b2) + 2ab(2a – b)

= (2a – b)( 4a2 + 2ab + b2 + 2ab) = (2a – b)(2a + b)2

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

giúp em với ạ TvT em đang cần gấp lắm huhu

b)

Sửa đề: \(125a^3+75a^2+15a+1\)

Ta có: \(125a^3+75a^2+15a+1\)

\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)

\(=\left(5a+1\right)^3\)

c) Ta có: \(64-96a+48a^2-8a^3\)

\(=-\left(8a^3-48a^2+96a-64\right)\)

\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)

\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)

\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)

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

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

Bài 1 :

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

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)

Tick đúng nha 

X^2-6+8

\(2x+8=2\left(x+4\right)\)

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

2(x+4)

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

\(\text{ 2x² + 6x - 8 }\)

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