\(e)x^4-2x^4+x^2 =x^2.x^2-2x.x^2+x^2+1 =(x^2)(x^2-2x+1) =x^2(x-1)^2 \)

\(f)27y^3-x^3 =(3y)^3-x^3 =(3y-3)(9y^2+3xy+x^2)\)

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

\(=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x\cdot\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}+\dfrac{1.\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

b)\(\dfrac{3\left(x+1\right)^2}{x^3-1}-\dfrac{1-x}{x^2+x+1}+\dfrac{3}{1-x}\\ =\dfrac{3\left(x+1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{3x^2+6x+3+x^2-2x+1-3x^2+3x+3}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{x^2+x+12}{x^3-1}\)

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

\(a)2x^3y-8x^2y+8xy =2xy.x^2-8x^2y+2xy.4 =2xy(x^2-4x+4) =2xy(x-2)^2\)

\(b)3x(x+5)-2(5+x) =(x+5)(3x-2)\)

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

\(d)(2x-3)^2-25y^2 =(2x-3)^2-(5y)^2 =(2x-3-5y)(2x-3+5y)\)

\(1)4-x^2-2xy-y^2 =4-(x^2+2xy+y^2) =4-(x+y)^2 \)

\(2)25-x^2+2xy-y^2 =5^2-(x^2-2xy+y^2) =5^2-(x-y)^2 =(5-x-y)(5+x-y)\)

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

\(4)x^2+4xy-16+4y^2 =x^2+4xy+4y^2-16 =(x-4y)^2-4^2 =(x-4y-4)(x-4y+4)\)

\(a)x^2-xz-9y^2+3yz =(x^2-9y^2)-(xz-3yz) =(x+3y)(x+3y)-z(x-3y) =(x-3y)(x+3y-z)\)

\(b)12x^3+4x^2-27xy-9y =4x^2(3x+1)-9y(3x+1) =(3x+1)(4x^2-9y)\)

\(c)x^4-25x^2+20x-4 =x^4-25x^2+20x-2^2 =x^4-((5x)^2-20x+2^2) =(x^2)^2-(5x-2)^2 =(x^2-5x-2)(x^2+5x-2)\)

\(d)x^2y^2.(x^2-9)-x^2+9 =x^2y^2.(x^2-9)-(x^2-y) =(x^2-9)(x^2y^2-1)\)

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

\(64x^3-27y^3\\ =\left(4x\right)^3-\left(3y\right)^3\\ =\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

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

     \(x^4-4x^3+4x^2\)

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

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