Answer
$\frac{2}{l^2}+\frac{1}{l^3}-\frac{4}{3l^4}$
Work Step by Step
Splitting the fraction into a sum of fractions leaves $$=\frac{6l^2}{3l^4}+\frac{3l}{3l^4}-\frac{4}{3l^4}=\frac{6}{3} \cdot \frac{l^2}{l^4}+\frac{3}{3} \cdot \frac{l}{l^4}-\frac{4}{3l^4}=\frac{2}{l^{4-2}}+\frac{1}{l^{4-1}}-\frac{4}{3l^4}$$ $$=\frac{2}{l^2}+\frac{1}{l^3}-\frac{4}{3l^4}$$
