All the info I've managed to find only shows basic applications of the formula, in this case, I have the following Laplace Transform: $L [\sin3t-\cos3t]^3$
Does the exponential remain in the integer? Does the negative sign belongs to the cos3t and I have to include it in the formula? I'm sorry for the question but I have only seen so far Laplace Transforms of multiplications, in this case, would it look like this?
$$\int _0^{\infty }\:[(-\cos3t)(e^{-st})(\sin3t)dt]^3$$
It works like substituting things in functions - you take everything inside the brackets, and you put that in the appropriate place of the definition of the transform.
So just like if you have $f(x) = 7x^2 - e^x$, then $f(y + z) = 7(y + z)^2 - e^{y + z}$, then if your Laplace transform is defined as:
$$\mathcal{L}[f(t)] = \int_0^\infty f(t) e^{-st} dt$$
then when $f(t) = (\sin 3t - \cos t)^3$, you get
$$\mathcal{L}[(\sin 3t - \cos t)^3] = \int_0^\infty (\sin 3t - \cos t)^3 e^{-st} dt$$
AS a side note, your notation $\mathcal{L}[\sin 3t - \cos t]^3$ is slightly ambiguous - usually you would put the entire function being transformed inside the brackets, so it could be read as wanting to find the transform of $(\sin 3t - \cos t)^3$ as I wrote above, but it could also be asking to find the transform of $\sin 3t - \cos t$, and then take the cube of the result.