Bài 2.

\(a.-x^3+3x^2-3x+1=\left(-x+1\right)^3\)    \(b.x^3+x^2+\frac{x}{3}+\frac{1}{27}=\left(x+\frac{1}{3}\right)^3\)

\(c.8-12x+6x^2-x^3=\left(2-x\right)^3\)     \(d.x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

bài 3.

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

\(B=\left(x^3-6x^2+12x-8\right)-\left(x^3-x\right)+6x^2-18x=-5x-8\)

Bài 4.

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

\(b.\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)\)\(=12x-4=-10\Leftrightarrow x=-\frac{1}{2}\)

mình làm mẫu a;c thôi nhé 

a, \(\left(6x^3+3x^2+4x+2\right):\left(3x^2+2\right)\)

\(=\left[2x\left(3x^2+2\right)+3x^2+2\right]:\left(3x^2+2\right)=\frac{\left(2x+1\right)\left(3x^2+2\right)}{3x^2+2}=2x+1\)

c, \(\left(12x^4+4x^3+9x+3\right):\left(3x-2\right)=\left[3x\left(4x^3+3\right)+4x^3+3\right]:\left(3x-2\right)\)

\(=\frac{\left(3x+1\right)\left(4x^3+3\right)}{3x-2}\)bạn xem lại đề ý c nhé 

\(\frac{6x^3+2x^2+4x+2}{3x^2+2}\)

\(=\frac{6x^3+4x+2x^2+2}{3x^2+2}\)

\(=2x+\frac{3x^2+2-x^2}{3x^2+2}\)

\(=2x+1-\frac{x^2}{3x^2+2}\)

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