Answer
$$\dfrac{-3s^2 -128}{s(s^2+64)}$$
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
We are given that $f(t)=4 \sin^2 (4t)-3$
Now, $L[F(t)]= L[4 \sin^2 (4t)-3] \\= L[4 \sin^2 (4t)] -L[3] =4 L[\dfrac{1-\cos 8t}{2}-L[3] \\=- \dfrac{s}{s^2 +64}-\dfrac{2}{s}\\=\dfrac{-3s^2 -128}{s(s^2+64)}$
