Use Ito's Formula to prove following identity Again, I am not sure how the following works; Could someone please give me an almost stupidly detailed explanation of why/what is happening in the part below. First, the question itself;

Q. $B_t$ is a standard Brownian motion. Use Ito's formula to show the following identity holds,
$$\int_0^tB_s^3ds=\frac{1}{4}B_t^4-\frac{3}{2}\int_0^tB_s^2ds$$

Answer(partial)

A. By Ito's formula, one obtains $dB_s^4=4B_2^3dB_2+6B_s^2ds$ and which by integrating both sides ...etc

The rest is fine. I just don't know how "Ito's formula" gives "$dB_s^4=4B_2^3dB_2+6B_s^2ds$". I thught Ito's formula was the super tedious stochastic equation.
Briefly, it says for a Ito process $X_t$ which is defined $dX_t=\mu dt+\sigma dW_t$ with $X_0=x_0$, we have for some new Ito Process $Y_t$,
$$dY_t=[\frac{\partial f}{\partial t}+\mu \frac{\partial f}{\partial t}+\frac{\sigma^2}{2}\frac{\partial^2 f}{\partial t^2}]dt+[\sigma \frac{\partial f}{\partial x}]dW_t$$
I thought this differential equation was the "Ito's Formula". But how does this lead to the answer above??? I guess it's something very elementary but as a newbie, no, I don't see what is going on at all. Can someone elaborate on what the answer above has skipped and abbreviated? The bit "one obtains" as to how?
Thank you very much in advance for your help
 A: First you have to decide on the notation, first you denoted Brownian motion by $(B_t)$ and then by $(W_t)$. I will use the notation $(B_t)$ for a (standard) Brownian motion.
Ito's formula applies to Brownian motion and if you want to see that it is indeed an Ito process note that 
$$ B_t = \int_0^t dB_s$$ and to adjust it to the form that you have written note that
$$dB_t = 0 \cdot dt + 1 \cdot dB_t.$$
I would say that a it is worth to look at Ito's formula in the following form, for an Ito process $X_t$ if $f$ is twice continuously differentiable on $[0, \infty) \times \mathbb{R}$ then
$$ df(t, X_t) = \frac{\partial f}{\partial t}(t, X_t)dt +\frac{\partial f}{\partial x}(t, X_t)dX_t +\frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t,X_t) (dX_t)^2.$$
Now let $f(x, t)= x^4$ then $\frac{\partial f(x,t)}{\partial t} = 0$,$\frac{\partial f(x,t)}{\partial x} = 4x^3$, $\frac{\partial^2 f(x,t)}{\partial x^2} = 12x^2$, therefore by Ito's formula and having in mind Ito's table (that is, $(dt)^2 = dtdB_t = dB_tdt=0 \mbox{ and } (dB_t)^2 = dt$)
$$dB_t^4 = 0 dt + 4B_t^3 dB_t + \frac{1}{2}\cdot 12 B_t^2 (dB_t)^2 =4B_t^3 dB_t + 6 B_t^2dt.$$
