[tex]\displaystyle \frac{1}{t+p}+\frac{3tp}{t^3+p^3}=\frac{1}{t+p}+\frac{3pt}{(t+p)(t^2-pt+p^2)}=\frac{t^2-pt+p^2+3pt}{(t+p)(t^2-pt+p^2)}=\frac{t^2+2pt+p^2}{(t+p)(t^2-pt+p^2)}=\\\\\frac{(t+p)^2}{(t+p)(t^2-pt+p^2)}=\frac{t+p}{t^2-pt+p^2}[/tex]