Question

How do you simplify `\frac { 9x + 81} { 4x ^ { 3} } \cdot \frac { x } { x ^ { 2} - 81}`?

Answer 1

`(9)/((4x^2)(x-9))`

Explanation:

Let's use our "cross-simplify" method.
Let's factor if there is any factorable part there.
We see that `9x+81=9(x+9)` and `x^2-81=(x+9)(x-9)`
Therefore, we now have:
`(9(x+9))/(4x^3)*x/((x-9)(x+9))` We rewrite `4x^3` as `4x*x*x`
`(9(x+9))/(4x*x*x)*x/((x-9)(x+9))` Now, remember that `(a*b*c)/(d*e)*d/(b*b)=(a*cancelb*c)/(canceld*e)*canceld/(cancelb*b)`

Therefore,
`(9cancel(x+9))/(4cancelx*x*x)*cancelx/((x-9)cancel(x+9))` So we now have:`(9)/(4x^2)*1/((x-9))=>(9)/((4x^2)(x-9))`
That is our answer!