Answer
$(3m+n)(3m-n)$ or $9m^2-n^2$
Work Step by Step
Step 1. Replace division with multiplication and inverse the second rational expression, we have
$\frac{27m^3-n^3}{3m-n}\div\frac{9m^2+3mn+n^2}{9m^2-n^2}=\frac{27m^3-n^3}{3m-n}\times\frac{9m^2-n^2}{9m^2+3mn+n^2}$
Step 2. Use the formulas $a^2-b^2=(a+b)(a-b)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$, we have $9m^2-n^2=(3m+n)(3m-n)$ and $27m^3-n^3=(3m-n)(9m^2+3mn+n^2)$
Step 3. Thus, we have $\frac{27m^3-n^3}{3m-n}\div\frac{9m^2+3mn+n^2}{9m^2-n^2}=\frac{(3m-n)(9m^2+3mn+n^2)}{3m-n}\times\frac{(3m+n)(3m-n)}{9m^2+3mn+n^2}$
Step 4. Cancel common factors, we have $\frac{27m^3-n^3}{3m-n}\div\frac{9m^2+3mn+n^2}{9m^2-n^2}=\frac{(1)(1)}{1}\times\frac{(3m+n)(3m-n)}{1}=(3m+n)(3m-n)=9m^2-n^2$
