\(\dfrac{x^5+x+1}{x^3+x^2+x}=\dfrac{\left(x^5+x^3+x\right)-\left(x^3-1\right)}{x\left(x^2+x+1\right)}=\dfrac{x\left(x^4+x^2+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}=\dfrac{x\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}=\dfrac{\left(x^2+x+1\right)\left[x\left(x^2-x+1\right)-x+1\right]}{x\left(x^2+x+1\right)}=\dfrac{x\left(x^2-x+1\right)-x+1}{x}=\dfrac{x^3-x^2+x-x+1}{x}=\dfrac{x^3-x^2+1}{x}\)