Question

$$\frac{x^{2}-81}{x^{3}+18x^{2}+81x}$$

Answer

(x-9)/(x*(x+9))

Solution


Rewrite \({x}^{2}-81\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=9\).
\[\frac{{x}^{2}-{9}^{2}}{{x}^{3}+18{x}^{2}+81x}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\frac{(x+9)(x-9)}{{x}^{3}+18{x}^{2}+81x}\]
Factor out the common term \(x\).
\[\frac{(x+9)(x-9)}{x({x}^{2}+18x+81)}\]
Rewrite \({x}^{2}+18x+81\) in the form \({a}^{2}+2ab+{b}^{2}\), where \(a=x\) and \(b=9\).
\[\frac{(x+9)(x-9)}{x({x}^{2}+2(x)(9)+{9}^{2})}\]
Use Square of Sum: \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\).
\[\frac{(x+9)(x-9)}{x{(x+9)}^{2}}\]
Simplify.
\[\frac{x-9}{(x+9)x}\]
Regroup terms.
\[\frac{x-9}{x(x+9)}\]