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CCSS.Math:

we're asked to factor for TC to the third power minus five D to the third power so the first thing that might jump out at you is that five is a factor of both of these terms I could rewrite this as five times eight C to the third power minus five times D to the third power and so you could actually factor out a 5 here so factor out of five and so if you factor out a 5 you get 5 times 8 C to the third power minus D to the third power so as you see factoring it really is just undistributing the five reversing the distributive property and when you write it like this you might it might jump out at you that 8 is a perfect cube it's 2 to the 3rd power C to the 3rd 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 re-written as 2 C 2 C to the third power because this 2 to the third power times C to the third power 8 C to the third power and then minus D to the third power minus D to the 3rd power and so this gives us right over here a difference of cubes and you can actually factor out a different factor a difference of cubes and you may or may not know the pattern so if I have a to the third minus B the third this can be factored as a minus B times a squared plus a B 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 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 the sum of cubes is also factorable it's factorable as a plus B times a squared minus a B plus B so once again I won't go through the time 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 figure 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 to see our a is to C and our B is D so let me write this a is equal to 2c and our B our B is equal to D our B is equal to D we have minus B 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 five open parentheses let me give myself some space so it's going to factor out into a minus B so a is - C - C - B which is D so its factors out as the difference of the two things that I'm taking the cube of - C - D times x and now I have a squared a squared is - C squared - C squared is the same thing as 4 C squared a squared let me make that a squared is equal to 2 C the whole thing squared which is equal to 4 C squared so it's 4 C squared plus a B plus a times B so that's going to be 2 C times D so plus 2 C times D 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 it actually could get rid of one set of parentheses this can be factored as 5 times 2 C - D times 4 C squared + 2 C D plus d squared and we are done