We know that force (F) has dimensions of $$\mathrm{MLT}^{-2}$$, velocity (V) has dimensions of $$\mathrm{LT}^{-1}$$, and time (T) has dimensions of $$\mathrm{T}$$. To express density ($$\rho$$), which has dimensions of $$\mathrm{ML}^{-3}$$, in terms of F, V, and T, we need to find the exponents of F, V, and T.

Let the dimensions of density be expressed as:

$$\mathrm{[M]}^a \mathrm{[L]}^b \mathrm{[T]}^c$$

Substituting the dimensions of F, V, and T:

$$\mathrm{[F]}^a \mathrm{[V]}^b \mathrm{[T]}^c = (\mathrm{MLT}^{-2})^a (\mathrm{LT}^{-1})^b (\mathrm{T})^c$$

Since density has dimensions of $$\mathrm{ML}^{-3}$$, we can set up the following equations:

$$a = 1$$ (for the mass term M)

$$a + b = -3$$ (for the length term L)

$$-2a - b + c = 0$$ (for the time term T)

Solving these equations, we get:

$$a = 1$$

$$b = -4$$

$$c = -2$$

Thus, the dimensional formula of density in terms of F, V, and T is:

$$\mathrm{FV}^{-4} \mathrm{T}^{-2}$$