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Factoring the Sum or Difference of Cubes - Problem 3 3,745 views

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Factoring trinomials. So behind me I have an expression that we’re trying to factor; x to the ninth minus quantity 5 minus y to the third. At our disposal we have two different potential equations. We could either have the difference of squares or the difference of cubes. I know that because we’re subtracting so therefore it has to be a difference formula.

So looking at it, first 5 minus y to the third, there’s no way that can be written as a square so therefore I know I have to be dealing with a difference of cubes. So what I have to do is first go back to our difference of cubes formula. So I know that a³ minus b³ can be factored as a minus b, the first sign agrees, a² plus ab, the second sign is different plus b².

So the trick for this problem is actually figuring out what is a and what is b? So there’s really no reason to multiply this 5 minus y to the third out because we’re really looking for something cubed, so really what we can say is b is equal to 5 minus y. We’re dealing with b³, we already have something cubed so that something is our b.

The next part we need to look at is x to the ninth and we want to write that as something cubed as well. When we’re dealing with exponents, power to a power we multiply, so we want to figure out what we need to put as our power in order when we multiply to get 9. 3 times 3 is 9, so then our x³ is going to be our a term. From here we know what a is, this is a and we know what b is. We can just substitute that into our formula. So we start with a, this is actually not the cute part but I just did that to figure out what my a was. Okay, so a is x³, so we can just plug that in every time we see a. This is x³, a is x³, so x³ squared is going to be x to the sixth, then we have a plus x³ there.

We then minus b from our first term but b is just equal to 5 minus y, so substituting that in, 5 minus y, put that a little close together, 5 minus y, and then lastly plus 5 minus y quantity squared. Now, we could distribute all these out typically most teachers are perfectly fine leaving it in this form. You know, we could distribute our negative sign, multiply this out, FOIL this out, but what we’ve really done is the bulk of the work which is figuring out what our a and b is, figuring out the fact that we have a difference of cubes and then plugging our two a and b’s into the equation to factor.

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