Question

How can I simplify it
`\left((8x^3-1)/(\left(2-(1)/(x)\right)\left(x^2-9\right))\right)\cdot (\left(x^2+2x-15\ \right))/(4x^3+2x^2+x)`

Answer 1

`\cfrac{8x^3-1}{(2-(1)/(x))(x^2-9)}\cdot \cfrac{x^2+2x-15}{4x^3+2x^2+x}\implies \cfrac{2^3x^3-1^3}{(2-(1)/(x))(x^2-9)}\cdot \cfrac{(x-3)(x+5)}{x(4x^2+2x+1)} \\\\\\ \cfrac{\stackrel{ \textit{difference of cubes} }{(2x)^3-1^3}}{((2x-1)/(x))(\underset{ \textit{difference of squares} }{x^2-3^2})}\cdot \cfrac{(x-3)(x+5)}{x(4x^2+2x+1)}`

`\cfrac{(2x-1)(4x^2+2x+1)}{((2x-1)/(x))(x-3)(x+3)}\cdot \cfrac{(x-3)(x+5)}{x(4x^2+2x+1)}\implies \cfrac{(2x-1)}{((2x-1)/(x))(x+3)}\cdot \cfrac{(x+5)}{x} \\\\\\ \cfrac{(2x-1)}{ ~~ (((2x-1)(x+3))/(x)) ~~ }\cdot \cfrac{(x+5)}{x}\implies (2x-1)\cfrac{x}{(2x-1)(x+3)}\cdot \cfrac{(x+5)}{x} \\\\\\ \cfrac{(2x-1)x}{(2x-1)(x+3)}\cdot \cfrac{(x+5)}{x}\implies \cfrac{x+5}{x+3}`