 Hey everybody, welcome to Tutor Terrific. Today I'm going to do a factoring video for you, but it's not going to be a quadratic factoring video like my previous series of three. This is a factoring video for the sum or difference of two cubes. Yes, you can factor cubes in various ways, and this is one of them. So, the factoring method for a sum of two cubes or a difference of two cubes is often written in the following three ways, or one of the three ways. If we have a sum of two cubes, that means I have two things that have nice cube roots, we can factor them the following way. Two parentheses. The first parentheses, we would write the cube root of the first term, plus the cube root of the second term, see how there's no powers of three. Then we will multiply that by a trinomial, the first term's cube root squared minus the product of the cube root of the two terms, plus the square of the second term's cube root, okay? Notice how the middle sign changes. Second, we could do the difference of a cube and another cube. First, we will write the cube root of both terms, subtracted the second one from the first, then we multiply that by the cube root of the first one squared, just like before. This time, plus the product of the cube roots of the two terms, plus the square of the cube of the second term. Notice how the middle side stayed the same, the second sign changed, and the third sign was still positive. If we were to combine these two rules into one, we would get the following. A cube, plus or minus B times A squared minus or plus, showing the sign changed, product of AB, plus B squared. So the way to remember these signs, the word soap is often used. Soap stands for S same, O opposite, and AP stands for always positive. Same, opposite, always positive. So that's the acronym used to remember the signs. So let's practice using this with two simple examples here. First example, X to the 6 plus Y to the 18th. First thing you do is check, can I take the cube root of both of these terms? Yes and yes, you would divide the power by 3. So this would be equal to in the first parentheses, X, 6 divided by 3, X squared plus Y, what's 18 divided by 3? That would be 6. Then we multiply that by the square of the cube root of this term. So the square of this term, X squared, squared, would be X to the fourth, multiply those two exponents together. Then opposite sign minus the product of these two terms, X squared, Y to the sixth, plus the second term squared. That would be Y to the 12th. 6 times 2 is 12. So that's how that factors. Let's check out this next one. Okay, 8R cubed minus 27. Can I take the cube root of both terms? Absolutely. The cube root of 8R cubed is equal to 2R and the cube root of 27 is equal to 3. So this would be 2R minus 3. Okay, the cube root of both terms subtracted same sign. Then you take the square of the first term, that would be 4R squared. Then we will add opposite sign, the product of these two terms squared. So it would be, sorry, the product of those two terms by itself. That would be 6R, 2R times 3. Then 3 squared, positive. So that would be 9. Like so. So those are some simple examples. Now let's look at some more challenging ones. Alright, so let's check out this first example on this page. 125 minus 64 Y to the 6th. This might strike you as a little odd, but just check and make sure that both terms have cube roots that are perfect. 125. What's the cube root of 125? That's 5. 5 times 5 times 5 is 125. What about 64 and Y to the 6? 64 does have a cube root. I know it has a square root of 8, but it has a cube root also of 4. Y to the 6th cube root is Y squared. Alright, so this factors with the difference of cubes. So let's do it. Keep the same sign, write the cube root of each term with a minus sign in the middle. So that's 4Y squared. Okay, so we got the first part. Next, I square this. 25 opposite sign. Then we multiply these two terms together. You get 20Y squared. And then we will square the second term. Always positive. That would be 16Y to the 4th. Alright, now many of you may be wondering, can I ever factor the quadratic term in the second part of the sum of difference of two cubes factorization? No, you cannot. The reason you never will be able to is because there is no 2 in that factorization. That's normal quadratic factorization. You have to have a double there. And you do not in this case. So this cannot be factored with binormal means. So don't attempt to do it. It's not worth trying. Okay, next example. 27X to the 9th plus 1. 1? Really? Yes. Yes, really. Check. Make sure each one is a sum of two cubes. Is this a cube of something? 27 cube root? That's 3. So yes. And X to the 9th, the cube root of that is X to the 3rd, dividing by 3. And 1. Is that a cube root of anything? Yes. You can take the cube root of 1 and you get 1. So we do have two cubes. So this will equal the cube root of this, which is 3X to the 3rd plus 1. I know I'm repeating the 1 because 1 is its own cube root. Next we have 3X to the 3rd squared. So that's 9X to the 6th. Opposite sign now. Same sign, opposite sign. Always positive coming up. Multiply these two together. You just get 3X cubed. Always positive square the 1. You get 1. Okay, awesome. So you've seen some tricky examples. Now it's time to really up the ante here. I'm going to change the direction slightly to factor completely. The sum and difference of two cubes factorization will play a role here, but we are going to need to factor with different methods also. Because look here, 3S cubed minus 81V to the 6th. Things look okay as far as the S and V powers. You can divide those by 3. But the numbers, they don't have perfect cube roots. 3 does not have a nice cube root, neither does 81. So we've got a little bit of a problem here. What we will always try to do when we notice that that is occurring is we will try and factor out greatest common factor from the two terms. Now if you look at these two terms, if 81 is a multiple of 3, we can pull out 3. 81 is a multiple of 3. If you divide it by 3, you get 27. So awesome. I'm going to rewrite this as 3 times S cubed minus 27V to the 6th. Okay, that binomial in parentheses is a difference of two cubes. So I can factor it using the difference of two cubes factorization. So this equals 3 times the cube root of S cubed S. That's the cube root of 27V to the 6th, which is 3V to the 2nd. Okay, next, we're going to do the quadratic term. S squared plus always positive. Sorry, opposite, excuse me, opposite. S times 3V squared is just 3SV squared and always positive, the third sign, the square of 3V squared, which would be 9V to the 4th. There's a 3 in front because I first factored it out before I even did the summer difference of cubes factorization. Okay, and this last example, your final exam for x cubed y to the 7th plus 32y. This looks mind bogglingly not like a sum or difference of two cubes. I can't take the cube root of 4, can't take the cube root of y to the 7th, can't take the cube root of 32 or y. Oh my goodness. Well, first, you will then, if you can't do that, you will factor out the greatest common factor. What is the largest factor or all the factors that both terms share? Well, 32 could be divided by 4, so I could pull out a 4. In addition, both terms have one power of y at least, so I could pull that out. Let's write what we have now that we've done that. Okay, when I do that, I get inside just x cubed y to the 6th plus, now divide this by 4y, I just get 8. Ah, look at this. I could take the cube root of this. I know there's two variables multiplied together. I could still take the cube root of that and I could take the cube root of 8, which is 2. Perfect. So I'm going to do this now. So I'm going to apply the sum of two cubes factorization. What's the cube root of x cubed y to the 6th? Nothing but xy squared, dividing both those powers by 3. Then I'm going to do the opposite sign. Oh, no, I'm going to do the same sign because we're in the very first part, plus, and then the cube root of 8, 2. Next comes the 3 term quadratic. This term squared. So x squared y to the 4th. That's xy squared squared. Opposite sign minus the product of these two, 2x y squared. And then we're going to finish that off with the square of 2. So plus, always positive, plus 4. All right, again, do not try and factor these quadratic terms. They do not factor because of how they're set up. Thanks guys for watching. This has been factoring a summer difference of two cubes. This is Falconator, signing out.