Welcome back, everyone. In this video, we're going to continue talking about special product formulas, and in this video, we're going to cover some formulas involving cubes of polynomials rather than just squares. But the idea is the same. We're just going to be using these special products rather than having to multiply out the long way. Let's check it out.

So, basically, I'm just going to jump right into the formulas over here. We have, cubes of binomials, which is \(a + b^3\) or \(a - b^3\). Remember, we talked about squares of binomials. Now we just have cubes of binomials, and there are more terms. But, basically, the way that this works is that the signs go plus, plus, plus, and then for a negative, you're going to do negative, plus, and negative. So one way to remember this is that for plus, everything is positive, and for minus, basically, the signs alternate. It goes negative, positive, and then negative. Okay?

So, alright. So if I take a look at my example here, I have \(x - 3^3\). So which one is it? It's basically just the \(a - b^3\). That's sort of the pattern that it fits. So, in other words, this is like my a and that's my b. So, in other words, x is a and 3 is b.

So what does this work out to? Well, this works out to I'm just going to write this out. \(a^3 - 3a^2b + 3ab^2 - b^3\). The rest is, I have to figure out what each one of these things becomes now that I know that \(a\) is \(x\) and \(b\) is \(3\). So let's get started.

What is \(a^3\)? Well, \(a^3\) just becomes \(x^3\). Now, what's the middle terms? So what does the first or second term become? This becomes 3, and then \(a^2\). If \(a\) is \(x\), then \(a^2\) is \(x^2\), and then \(b\) is just \(3\).

And then what about the 3rd term? This becomes 3. A is \(x\), and then \(b\) is 3, so \(b^2\) just becomes \(9\). And then, minus \(b^3\), so that's \(3 \times 3 \times 3\), which is just \(27\). Alright.

So notice how fast this was. If we had to foil it out, it would have been a nightmare because we ended up with a 3 term, and then you got to multiply that again. It just would have been a disaster. So these formulas are, again, relatively speaking, pretty quick and straightforward. Alright.

So the last thing we just have to do is now simplify this. So the \(x^3\) now just goes straight down. And what does the second term become? I have a \(3 \times 3 \times x^2\), so this just becomes \(9x^2\). The middle third term actually goes to \(3 \times 27 \times\) just \(1\) power \(x\), so this is \(27x\), and then I have minus \(27\) over here. And that is the simplest I can make this. I can't combine any more terms or anything like that. Alright?

Alright. So this is my expression over here, and that's how you do the cubes of binomials.

Now with these types of formulas, almost always, or you're almost never going to have to memorize these. Your professors may actually give you something like a formula sheet. But in case they don't, I'm going to give you a good way to think about this, and sort of memorize them. We've already talked about the signs. If you actually notice what happens to the coefficients, the coefficients always go \(1, 3, 3,\) and \(1\). Right? So I have a \(1 a^3\), a \(3\), and then \(3\), and then a \(1 b^3\) over here.

So that's how the coefficients always go. There's also a pattern with the powers of \(a\) and the powers of \(b\). The powers of \(a\) decrease as you go from left to right. So we have an \(a^3\), then an \(a^2\), then an \(a\) to the one power. And then if you have no \(a\) over here, it's basically like you have \(a\) to the zero power. So, in other words, the powers of \(a\) always go from \(3\) to \(1\) to \(0\), and for \(b\), it actually goes the opposite. The powers of \(b\) actually increase. Notice how here we have a \(b\) to the zero power, then \(b\) to the \(1\), then \(b\) to the \(2\), and \(b\) to the \(3\). So this goes \(0, 1, 2,\) and \(3\). So, basically, \(a\) goes down and \(b\) goes up, and the coefficients are always \(1, 3, 3,\) and \(1\).