Question

\(\frac{1}{x^4-x}:\frac{x+1}{x^3+x^2+x}\)

Answer 1

\(\frac{1}{x^4-x}:\frac{x+1}{x^3+x^2+x}\)

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

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

= \(\frac{1}{x-1}.\frac{1}{x+1}\)

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