Algebraic simplification in calculus: how to get such a result? I'm almost in the very beginning of my math journey. Please, have patience because I'm not a math guy (which I'm really willing to change now).
I'm trying to learn calculus for my math class at university. And I'm watching the 3Blue1Brown videos. Were are trying to find the derivative of $s(t) = t^3$.  So my question is: how did he simplify this part:
$$\frac{ds}{dt} = \frac{(2+dt)^3 - (2)^3}{dt}$$
to get
$$ \frac{2^3 + 3(2)^2dt + 3(2)dt^2 + dt^3 - 2^3}{dt}$$
At first, I thought it would be binomial to the power of 3 rule, then monomial to the power of 3 simplification, and something related with the subtraction. But alas my knowledge is not enough to comprehend it.
UPD: I thought that the numerator had to be:

*

*$$(2 + dt)(2 + dt)(2 + dt) - (2)(2)(2)$$

*$$2(2 + dt) + dt(2 + dt)(2 + dt) - (2)(2)(2)$$
UPD: And this is where I stumbled upon I guess. I had to apply the binomial formula when binominal is being brought to the power of 3, right? Then $$(2+dt)^3$$ would look like $$a^3 - 3ab(a - b) - b^3$$
I'm really having hard time trying to understand it. What rules apply to this simplification? What topics do I need to know before trying to solve this?
 A: I think the issue here is actually 3B1B's presentation, which is surprising, as he's usually really good with explanations. I'm not sure why he decided to computer at a specific value of $t$. That always seems to muddle things up. (Also I agree with Dave Renfro in the comments, that using $dt^2$ is kind of an abuse of notation).
Instead of $t=2$, let's consider an indeterminate $t$. We still basically start the same way:
$$\frac{ds}{dt} = \frac{s(t + dt) - s(t)}{dt} = \frac{(t+dt)^3-t^3}{dt}$$
Then we expand the polynomial:
$$\frac{ds}{dt} = \frac{ \color{blue}{t^3 + 3t^2 dt + 3t(dt)^2 + (dt)^3} - t^3}{dt} = \frac{3t^2dt+3t(dt)^2 + (dt)^3}{dt}$$
Notice that the portion in $\color{blue}{\text{blue}}$ is the same as the $(2+dt)(2+dt)(2+dt)$, just using $t$ instead of the specific value $2$. Now we can factor out $dt$ on the top of the RHS and divide, giving:
$$\frac{ds}{dt} = 3t^2 + 3tdt + (dt)^2$$
If we now decrease $dt$ so that $dt \to 0$, we finally simply get:
$$\frac{ds}{dt} = 3t^2$$
For $t=2$, we can then calculate $ds/dt=3 \cdot 2^2 = 12$.
The huge thing to understand is that $dt$ is a "quantity" we can manipulate algebraically, but doesn't properly "exist" in the way that, say, $t^3$ does. After any algebra is done to it, it must be moved down to $0$. This is called the method of infinitesimals, and I think it's really useful to look at calculus from this perspective, but do understand that it's a non-standard way of looking at things, and usually isn't taught, which I feel is a huge shame.
