Answer
$L[ e^t-te^{-2t}]=\dfrac{1}{(s-1)}-\dfrac{1}{(s+2)^2}$
Work Step by Step
The Laplace transform of function $t^3$ is given as:
$L(e^t)=\dfrac{1}{s-1}$ and $L(t)=\dfrac{1}{s^2}$
The first shifting Theorem for $a=- 2$ can be expressed as:
$L[ e^t-te^{-2t}]=\dfrac{1}{(s-1)}-\dfrac{1}{(s+2)^2}$