Найдите значение выражения (n^2+n)/(n^3-8)-(n+4)/(8-n^3)

[latex] \frac{n^2+n}{n^3-8}- \frac{n+4}{8-n^3}= \frac{n^2+n}{n^3-8}+ \frac{n+4}{n^3-8}= \frac{n^2+n+n+4}{n^3-8}=\\\\= \frac{n^2+2n+4}{(n-2)(n^2+2n+4)}= \frac{1}{n-2} [/latex]

n^2+n+n+4\(n-2)(n^2+2n+4)=n^2+2n+4\(n-2)(n^2+2n+4)=1/n-2