Tip:
Highlight text to annotate it
X
Sum and difference of cubes. This is a method for factoring polynomials, and what is factoring?
That means breaking something up into things that multiply to give you the thing that you started out with.
So when do you use this particular method for factoring polynomials?
Well, first of all you have to have a binomial; so that means two things. Two terms that are separated by a plus sign.
Something here plus something, or minus because it's sum or difference.
And second, both terms have to be perfect cubes.
Here are some examples. Each one of these are binomials. They all have to two things:
one, two, one, two, and all of this is one, and all of this is two. They're all separated by pluses or minuses,
and they're all perfect cubes as well.
So 8, for example is 2-cubed. x-cubed is obviously a cube, a-cubed is a cube. 1 is 1-cubed; 1 to the power of anything is 1.
125 is 5-cubed, and 1000 is 10-cubed.
And it even works with the a to the 6th and a to the 9th power, because a to the 6th power is a squared and then cubed.
So it'll work anytime you have a letter with an exponent that is a multiple of 3. 'b' to the 9 is b-cubed and then cubed. It's kind of a double cube.
Now here are the two formulas for the sum of cubes and difference of cubes methods for factoring.
If you're required to memorize these, I'll give you a few hints.
The first thing you might notice if you look at these two formulas is that both of them have the letters in the same place.
The x, y, x-squared, xy, and y-squared is the same in both of them.
What's different is the signs in between them, the pluses and minuses.
So the way I remember this is that the first sign is always the same as what you started out with, so this one is the same as this one over here.
If you're using the sum of cubes method, the first one will be a sum.
This one is always the opposite here, and the last one is just always plus. So you'll see it's the same down here: same, opposite, plus.
Same, opposite, plus! That's how you can remember the signs, and then you just have to remember it's always x, y, and then x-squared, xy, and y-squared.
It kinda makes it symmetrical.
So I'll show you my method for doing the sum of cubes, and it also works with the difference of cubes factoring.
The first thing is figure out all the things that go into the formula, so what are the x-cubed, y-cubed, the x, the y, the x-squared, the x times y, and the y-squared that goes in.
Figure them all out separately, then substitute them in, then you're done... usually... but you should check that it's factored completely.
That should always be the last step in a factoring question, because they always want the answer to be factored completely.
There might be a way you can factor it more with a different method. It's possible.
Now here's an example. We have to factor the polynomial 8 a cubed plus 27.
And you can see this is a binomial; there are two things, they're separated by a plus, and they are both perfect cubes.
So I'm just going to name the 8 a cubed the "x-cubed" from the formula, and the 27 will be the "y-cubed".
And I'll write the formula over here, just as a reference:
x-cubed plus y-cubed equals x plus y because it's the same, and x squared minus xy - minus is opposite - plus y squared. Same opposite plus.
So now you think about which values are the x squared, the xy, and the y squared, and so on.
Actually, start with the x-cubed. So the x-cubed in the formula, I already said is turning into 8 a cubed, and y cubed is turning into 27.
So the first part was already done for us. It was already in the question.
Now, what is x? We know x-cubed is 8 a cubed, so what do you cube to get 8 a cubed.
Well, 2 cubed gives you 8, so we'll have 2, and a cubed is just a-cubed. So x should be 2a, and just to show you more of where that comes from.
You know that if you cube 2a - the whole thing - you get 8 a cubed, because you cube the 2 and the a separately.
Now y. So we know y-cubed is 27, so that means y has to be 3. Because 3 times 3 times 3 is 27.
Now what's x-squared? It'll be 2a all squared, so that means we square the 2, which is 4, and then a squared is just a-squared.
And then what's y-squared, it'll be 3 squared, 3 times 3 is 9.
Now, the last thing is just find x times y. So x was 2a, y was 3, so 2a times 3 is 6a. Now we know everything.
So here's everything we just figured out, and all that's left is to substitute it all into the formula.
So we start with the x cubed, which we turned into 8 a cubed, and y cubed was 27, so 8 a cubed plus 27 equals...
x was 2a, y becomes 3, x squared becomes 4 a squared, minus x times y over here, that's 6a, plus 9.
And that's it! That's the answer!
You can check to make sure there aren't ways you can factor this, just to make sure that it's fully factored.
But there aren't. So neither this part in front, nor this part can be factored any more.
Here's another example of factoring; we're asked to factor 64 x-cubed minus 1. Now first of all, what method do we use?
There are two things, and hopefully you can see they are both perfect cubes, because 64 is 4 cubed, 1 is just 1 cubed, and x cubed is obviously a cube,
and it's a difference because there's a minus, so we're going to use the difference of cubes method.
So I'm gonna write that down... x-cubed minus y-cubed equals - and remember it's the same sign coming next, x minus y,
then it's the opposite sign - x-squared plus xy - the last sign is always plus - so plus y-squared.
Now, we start by figuring out the first two things there, the x-cubed and the y-cubed,
so x cubed - and I always put these little arrows because x-cubed isn't equal to 64 x-cubed, but that's what I'm substituting for it.
Now y-cubed is going to turn into 1, because that's the second number there. I don't have to put the minus because it's part of the formula.
So next I figure out x, that's the next one in line anyway. So what do you have to cube to get 64 x-cubed?
And the answer is 64 is 4 cubed, and you have to put that x in there. If you cube this whole thing you get 64 x cubed, and y is just going to be 1.
Now, x squared. Just take what we just figured out - the 4x - and square it, so it'll be 16 - that's 4 times 4 - x squared.
And y-squared is going to be 1 squared, which is just 1.
And the last thing we need to figure out - I'll put it over here - is x times y. From right here. So that is turning into 4x times 1, so it'll just be 4x.
Now, I'm going to substitute it into the formula down here. So we start out with x-cubed.
turns into 64 x-cubed and minus 1 equals 4x - that's the x - minus 1 - so look up here, that's what I'm substituting from -
x-squared is 16 x-squared, plus xy is 4x, plus y-squared was just 1 again.
So that's it. That's that thing factored because you can't factor it anymore.
Here's yet another example of factoring using sum or difference of cubes. This one doesn't look like you can use the sum or difference of cubes.
There are two things here, but 16 isn't a perfect cube and neither is 250, so what you have to do is factor it using a different method first.
You may notice that there are common factors in each one of these two terms, so you can factor those out.
A common factor between 16 and 250 is 2. So I can put that in front, and also
you have 'a' terms in both of them, so I can put down the lowest power of 'a' that you see, a to the 4.
So after that comes - you divide each of them by the numbers in front, so 16 divided by 2 is 8. The a to the 4 here just disappears and becomes a 1 and your left with just b to the 6.
plus, and then I divide the 250 by 2 as well, and that gives you 125,
and a to the 10 I'm gonna divide by a to the 4, and you end up subtracting the two exponents
so you have a to the 6.
So we've already factored this using one method, but hopefully you can see we're not done yet because
what's inside the bracket here is now a sum of two cubes.
So I can factor this more now. So I'll write down the formula over here really quickly,
x cubed plus y cubed equals x plus y times x squared minus x times y plus y squared.
Now, I take the two things here, and substitute them in, so the x cubed from the formula is turning into 8 b to the 6th. And the y cubed is turning into 125 a to the 6th.
Now, x is just the cube root of what we started out with here so that's gonna be 2 b squared.
Now can you see why that is?
If I take two b squared and cube it - all in brackets - you get back to where we started here;
the 2 cubed becomes an 8, the b-squared cubed becomes 6 because the 2 and the 3 are multiplied.
The same thing for 'y'. So y is going to be 125 so we have 5 cubed gives you 125 and a-squared cubed gives you a to the 6th.
Next, x-squared is going to be what x was and just square it, so 4 b to the 4.
y squared is 25 a to the 4.
And the last thing is just x times y.
So that's going to turn into 10 a squared b squared.
Now, all that's left is to substitute that back into the formula. And I'm not gonna write out the x-cubed and y-cubed yet,
I'm just gonna put the final answer, so we still have the 2 a to the 4 in front so don't forget that that's there.
And instead of the 8 b to the 6th plus 125 a to the 6th, we have this part...
x plus y times x squared minus x times y plus y squared.
So I'll substitute that in all at once, so it'll be 2 b squared plus 5 a squared...
and then you have 4 b to the 4th minus 10 a squared b squared plus 25 a to the 4th.
And that's it! That will be the final answer. You can't factor that anymore.
Thank goodness!
Now it looks complicated, but really it's not so much.
One of the hardest parts is just identifying that you have to use this method.
The sum of cubes. Then it's just step by step - figure out what all the things in the formula are and just substitute them in.
And the answer is big and long and ugly, but it still works.
Subtitles by the Amara.org community