# Factoring difference of cubes

CCSS Math: HSA.SSE.A.2

## Video transcript

We're asked to factor 40c to the third power minus 5d to the third power. So the first thing that might jump out at you is that 5 is a factor of both of these terms. I could rewrite this as 5 times 8c to the third power minus 5 times d to the third power. And so you could actually factor out a 5 here, so factor out a 5. And so if you factor out a 5, you get 5 times 8c to the third power minus d to the third power. So as you see, factoring, it really is just undistributing the 5, reversing the distributive property. And when you write it like this, it might jump out at you that 8 is a perfect cube. It's 2 to the third power. c to the third power is obviously c to the third power. And then you have d to the third power. So this right here is a difference of cubes. And actually, let me write that explicitly because 8 is the same thing as 2 to the third power. So you can write this as-- let me write the 5 out front-- 5 times-- this term right over here can be rewritten as 2c to the third power because it's 2 to the third power times c to the third power-- 8c to the third power-- And then, minus d to the third power. And so this gives us, right over here, a difference of cubes. And you can actually factor a difference of cubes. And you may or may not know the pattern. So if I have a to the third minus b to the third, this can be factored as a minus b times a squared plus ab plus b squared. And if you don't believe me, I encourage you to multiply this out, and you will get a to the third minus b to the third. You get a bunch of terms that cancel out, so you're only left with two terms. And even though it's not applicable here, it's also good to know that the sum of cubes is also factorable. It's factorable as a plus b times a squared minus ab plus b squared. So once again, I won't go through the time of multiplying this out, but I encourage you to do so. It just takes a little bit of polynomial multiplication. And you'll be able to prove to yourself that this is indeed the case. Now, assuming that this is the case, we can just do a little bit of pattern matching. Because in this case, our a is 2c, and our b is d. So let me write this. a is equal to 2c, and our b is equal to d. We have minus b to the third and minus d to the third, so b and d must be the same thing. So this part inside must factor out to-- let me write my 5, open parentheses. Let me give myself some space. So it's going to factor out into a minus b. So a is 2c minus b, which is d. So it factors out as the difference of the two things that I'm taking the cube of. 2c minus d times-- and now, I have a squared is 2c squared. 2c squared is the same thing as 4c squared. Let me make that. a squared is equal to 2c-- the whole thing squared, which is equal to 4c squared. So it's 4c squared plus a times b. So that's going to be 2c times d, so plus 2c times d. And then finally plus b squared, and in our case, b is d. So you get plus d squared. And you're done. We have factored it out. And actually, you could get rid of one set of parentheses. This can be factored as 5 times 2c minus d times 4c squared plus 2cd plus d squared. And we are done.