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Find $L^{-1}\left\{\frac{12}{s^{3}-8}\right\}$

$\frac{12}{s^{3}-8}=\frac{12}{(s-2)\left(s^{2}+2 s+4\right)}=\frac{A}{s-2}+\frac{B S+c}{s^{2}+2 s+4}$

$=\frac{A\left(s^{2}+2 s+4\right)}{(s-2)\left(s^{2}+2 s+1\right)} +\frac {{Bs + c}}{s^{2}+2 s+4}\frac {(s-2)}{(s-2)}$

$12=A\left(s^{2}+2 s+4\right)+(B s+c)(s-2)$

$S=2 \quad \longrightarrow \quad 12=A(4+4+4)+0 \rightarrow A=\frac{12}{12}=1 \quad A=1$

$S=0 \quad \longrightarrow 12=A(0+0+4)+(c)(-2) \rightarrow 12=4-2c$

$-2 c=8 \rightarrow c=-4$

$S=1 \quad \longrightarrow 12=1(1+2+4)+(B-4)(1-2)$

$12=7+(B-4)(-1)=7-B+4$

$\rightarrow B=7+4-12$

$B=-1$

$\mathcal{L}^{-1}\left\{\frac{12}{s^{3}-8}\right\}=\mathcal{L}^{-1}\left\{\frac{12}{(s-2)\left(s^{2}+2 s+4\right)}\right\}=\mathcal{L}^{-1}\left\{\frac{A}{(s-2)}+\frac{B s+c}{s^{2}+2 s+n}\right\}$

$=\mathcal L^{-1}\left\{\frac{1}{s-2}+\frac{-s-4}{s^{2}+2 s+4}\right\}$

$=e^{2 t}-\mathcal {L}^{-1}\left\{\frac{S+4}{s^{2}+2 s+4}\right\}$

$S^2+2s +4=s^{2}+2 s+1-1+4=(s+1)^{2} +3$

$=e^{2 t}-\mathcal {L}^{-1}\left\{\frac{s+4}{(s+1)^{2}+3}\right\}$

$=e^{2 t}-e^{-t} \mathcal {L}^{-1}\left\{\frac{s+3}{s^{2}+3}\right\}$

$=e^{2 t}-e^{-t}\left[\cos (\sqrt{3} t)+\frac{3}{\sqrt{3}} \sin (\sqrt{3}t)]\right.$