How to solve this laplace transform

Find the inverse laplace transform of the given function:
F(s)= (3e^-6s)/(s^2-9)
uc(t) denotes the heaviside function, which is 0 for t=c

a.u6(t)sinh3(t-6)
b.sinh((t-6)/3)
c.u6(t)cos((t-3)/6)
d.u6((t)cosh3(t-6)
e.u3(t)sin6(t-3)

F(s) = (3 e^(-6s) ) / (s^2 - 9)

Use the following fact:

L⁻¹ { e^(-cs) } F(s) } = u(t - c) f(t - c)

In your case we have c = 6, so the problem is reduced to:

u(t - 6) L⁻¹ { 3 / (s^2 - 9) }
u₆(t) L⁻¹ { 3 / (s^2 - 9) }

But now we can use the following fact:

L⁻¹ { a / (s^2 - a^2) } = sinh(at)

In your case we have exactly a = 3, so:

u₆(t) sinh(3t)

But remember that we actually need f(t - c) for c = 6, so the final answer is:

u₆(t) sinh(3(t - 6))

Thus the correct answer is A.

Done!