Respuesta :

LammettHash

Recall that for integers n > -1,

[tex]L_s\left\{t^n\right\} = \dfrac{n!}{s^{n+1}}[/tex]

Then

[tex]\displaystyle L^{-1}_t\left\{\frac2{s^3}\right\} = L^{-1}_t\left\{\frac{2!}{s^{2+1}}\right\} = \boxed{t^2}[/tex]

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