a) `64-96a+48a^2-8a^3`

`=-(8a^3-48a^2+96a-64)`

`=-[(2a)^3 - 3.(2a)^2 .4 + 3.2a.4^2 - 4^3]`

`=-(2a-4)^3`

b) `-m^3n^6-8`

`=-(m^3n^6+8)`

`=-[(mn^2)^3+2^3]`

`=-(mn^2+2)(m^2n^4-2mn^2+4)`.

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: \(64-96a+48a^2-8a^3\)

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

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

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

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

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

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

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

6x^3-x^2-486x+81

=x²(6x-1)-81.(6x-1)

=(x²-81)(6x-1)

=(x²-9²)(6x-1)

=(x-9)(x+9)(6x-1)

\(-2x^3+x^2+12\)

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

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

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

\(8x^4+81\)

\(=8x^4+2\cdot2\sqrt{2}\cdot x^2\cdot9+81-36\sqrt{2}\cdot x^2\)

\(=\left(2\sqrt{2}x^2+9\right)^2-\left(6\sqrt[4]{2}\cdot x\right)^2\)

\(=\left(2\sqrt{2}\cdot x^2-6\sqrt[4]{2}\cdot x+9\right)\left(2\sqrt{2}\cdot x^2+6\sqrt[4]{2}\cdot x+9\right)\)

a: =(x+6)(x-1)

n: \(=4x^4+36x^2+81-36x^2\)

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

a. \(-x^3-6x^2+6x+1=-x^3+x^2-7x^2+7x-x+1=\left(1-x\right)\left(x^2+7x+1\right)\)

b. \(x^4-4x^2+4x-1=x^4-1-4x\left(x-1\right)=\left(x-1\right)\left[\left(x+1\right)\left(x^2+1\right)-4x\right]\)

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

c. \(6x^3-x^2-486x+81=6x^3-54x^2+53x^2-477x-9x+81=\left(x-9\right)\left(6x^2+53x-9\right)\)

\(=\left(x-9\right)\left(x+9\right)\left(6x-1\right)\)

d. \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)=x^2\left(x^2+8x+16\right)-x^2-8x-16-x^2+1\)

\(=x^4+8x^3+14x^2-8x-15=x^4+5x^3+3x^3+15x^2-x^2-5x-3x-15\)

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

Để phân tích nhân tử các dạng này, em cần nhẩm được nghiệm để biết đc nhân tử chung là gì, sau đó tách để xuất hiện nhân tử chung đó. CHÚC EM HỌC TỐT :)) 

\(x^3-4x^2-8x+8\)

\(=x^3+2x^2-6x^2-12x+4x+8\)

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

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