## Elementary Algebra

$(3n-5)(9n^2+15n+25)$
Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the difference of $2$ cubes, the factored form of the given expression, $27n^3-125 ,$ is \begin{array}{l}\require{cancel} (3n-5)[(3n)^2+3n(5)+(5)^2] \\\\= (3n-5)(9n^2+15n+25) .\end{array}