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

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

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

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

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

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

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

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

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

$$\rightarrow B=7+4-12$$

$$B=-1$$

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

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

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

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

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

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

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

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