1/(s³+1)=⅓1/(s+1)+⅓(2-s)/(s²-s+1) in partial fractions.
s²-s+1=(s-½)²+¾.
ℒ⁻¹{1/(s+1)}=e⁻ᵗ, so ℒ⁻¹{⅓1/(s+1)}=⅓ℒ⁻¹{1/(s+1)}=⅓e⁻ᵗ.
2-s=3/2-(s-½), so ⅓(2-s)/(s²-s+1)=
⅓[(3/2-(s-½))/(s²-s+1)]=
½1/((s-½)²+¾)-⅓(s-½)/((s-½)²+¾)=
(√3/3)(√3/2)/((s-½)²+¾)-⅓(s-½)/((s-½)²+¾).
ℒ⁻¹{(√3/3)(√3/2)/((s-½)²+¾)-⅓(s-½)/((s-½)²+¾)}=
(√3/3)e⁰˙⁵ᵗsin(t√3/2)-⅓e⁰˙⁵ᵗcos(t√3/2).
The complete inverse is:
ℒ⁻¹{1/(s³+1)}=⅓e⁻ᵗ+(√3/3)e⁰˙⁵ᵗsin(t√3/2)-⅓e⁰˙⁵ᵗcos(t√3/2).