In Exercises 69–80, factor completely.x³ − y³ − x + y

Question

Accepted Answer

Hello, today we're gonna be completely factoring the given expression. So what we are given is A to the sixth B cubed plus A, a cubed minus 27 A cubed B cubed minus 216. Now, what we have here is a very complicated polynomial of four terms. And whenever you have a polynomial of four terms, what you want to go ahead and try doing is factoring by grouping. So we're going to group up the first two terms and the last two terms together. And we're gonna go through each group and factor out a greatest common factor. So let's go ahead and take a look at the first group. We have A to the sixth B cubed plus eight A cubed. In these two terms, there's at least an A cubed in both terms. So we can go ahead and factor out a cubed from the first group and taking a look at the second group, we have negative 27 negative 216. Both of these numbers share a common multiple of 27. But after we factor out the greatest common factor, we want this leading term to be positive. So we're gonna go ahead and factor out a negative 27 from the second group. So if we factor out these common factors, what we are left with is a cubed times the quantity A cubed B cubed plus eight minus 27 times a cubed B cubed minus or I'm sorry, plus because we factor out a negative number plus eight. Now notice that we have two instances of A Q B B cubed plus eight. We can go ahead and combine this into a single group by adding the coefficients together. And if we add the coefficients together, what we get is a cubed minus 27 times the quantity A cubed B cubed plus eight. Now both of these groups can be factored even further. And for now, I'm gonna go ahead and separate the groups. I'm gonna write the first group as number one. I'm gonna write the second group as number two. Now let's go ahead and factor each group individually. If we take a look at group number one, we have a cubed minus 27. Now a cubed is just a times itself three times. So it's a perfect cube. Furthermore, 27 can be rewritten as three cubed. So if we rewrite the first group as a cubed minus three cubed, what we have here is a difference of cubes and in order factor difference of cubes that states that if we have the quantity A cube to minus B cubed. This can be rewritten as A minus B times the quantity A squared plus A times B plus B squared. Here A is just going to equal to A and B is going to equal to three. So if we factor the group using the difference of cubes, we can rewrite this as A minus B which is going to be a minus three times the quantity A squared plus A times B and A times three is going to give us three, A plus B squared, three squared is going to give us nine. So this is going to be the factored form of the first group. Now let's go ahead and take a look at the second group. The second group is a cubed B cubed plus eight. A cubed B cubed is just A times B times itself three times. So we can rewrite this first term as A B cubed and eight is a perfect cube because eight is two times two times two, which itself is too cubed. So we can rewrite the second term to be too cubed. Now, what we have here is a sum of cubes and if we want a factor a sum of cubes, what that states is that if we have the quantity A cubed plus B cubed, this can be factored as A plus B times the quantity A squared minus eight times B plus B squared. Here we're going to let A equal to A B and we're going to let B equal to two. And if we plug this into the sum of cube terms, what we get is A plus B, which is going to be A B plus two times the quantity A squared, which is going to be A squared, B squared minus A times B which is going to be negative two A B plus B squared, which is going to be two squared, which is going to simplify to give us four. So this is going to be the factored form of the second group. And now if we combine all of our factors together, the factored form of the original expression is going to be a minus three times the quantity A squared plus three A plus nine times the quantity A B plus two times A squared B squared minus two A B plus four. And this is going to be the completely factored version of the given expression. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 69–80, factor completely.x³ − y³ − x + y

Key Concepts

Factoring Polynomials

Difference of Cubes

Grouping Method

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