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Factoring sum of cubes

Sal factors 27x^6+125 as (3x^2+5)(9x^4-15x^2+25) using a special product form for a sum of cubes. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • blobby green style avatar for user bossricky8
    Can anyone tell me where to start and end on videos that have to do with Algebra II.
    And if there is more than one place to see the Algebra II topics, please list ALL of them.
    I'm kind of stuck on topics for the final exam this semester in this class and I can't organize where to watch these videos, so I can fully understand all of Algebra II. I know there is an Algebra I playlist and companion playlists, but I need to know where to get ALL of the Algebra II topics.
    Thanks for your help!!!
    (36 votes)
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  • blobby green style avatar for user staples1027easytech
    what if you have a problem that looks like x^3+8 ?
    (7 votes)
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  • blobby green style avatar for user kelseys103
    Thank you! My Algebra 2 teacher is terrible...you have saved my GPA!

    Now for my question:

    I'm really confused on how you would rewrite the 27x^6. I understand it if there is a number in front and it has an exponent that can be simplified but (for instance) if you had x^3. You don't have a number in front and the exponent can not be simplified. How would I rewrite something like this?

    I could really use your help.

    ~Kelsey
    (7 votes)
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  • blobby green style avatar for user Dennis Xavi Nunez
    Why is there a - in front of the parentheses of (3x^2+5)
    I believe thats a mistake
    (4 votes)
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  • piceratops ultimate style avatar for user IcewolfAndrew
    Why is there a negative sign at ? There shouldn't be a negative sign there.
    (4 votes)
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  • blobby green style avatar for user kaylila.mae.faith
    how would you solve 64^2-y^3.

    My teacher never really went in depth with this process and i'm confused if this can even be solved with the perfect cube patterns?
    (2 votes)
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    • leaf green style avatar for user ArDeeJ
      What do you mean by solving? What do you have to do? I'm going to assume you have to factor it.

      As it happens, 64 = 4^3, that's where you get the cube from.
      64^2 = (4^3)^2 = (4^2)^3 = 16^3.

      Substituting that gives 64^2 - y^3 = 16^3 - y^3 = (16 - y)(16^2 + 16y + y^2).
      (6 votes)
  • piceratops seed style avatar for user Larissa.Amato
    Ive watched this a few times and completely understand
    I try and apply it to my math homework but keep getting the wrong answer.
    The problem is
    x+125x^4
    I factor it to x(1+125x^3)
    the directions says to use the sum of two cubes but I don't see two cubes and I keep getting the wrong answers
    please help! its an online class and I have yet to have my teacher respond to my emailed questions.
    (2 votes)
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    • orange juice squid orange style avatar for user N Peterson
      You're partway there.
      1+125x^3 can be factored more.
      1 is a perfect cube (1 * 1 * 1=1), and so is 125x^3 (5x * 5x * 5x=125x^3)
      The formula for the sum of cubes (a^3+b^3) is
      a^3+b^3=(a+b)(a^2-ab+b^2)
      Plugging in your numbers, we get
      (1+5x)(1+5x+25x^2)
      Remember the x you factored out.
      (x)(5x+1)(25x^2+5x+1)
      FYI:
      a^3-b^3=(a-b)(a^2+ab+b^2)
      a^3+b^3=(a+b)(a^2-ab+b^2)

      Hope I could help
      (2 votes)
  • purple pi teal style avatar for user SnowRabbit
    Any tips on remembering where exactly the minus sign goes in the sum of cubes formula? Same with difference of cubes..
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      First, the 2 patterns are very similar. They are identical except for 2 signs.
      a^3+b^3 = (a+b)(a^2-ab+b^2)
      a^3-b^3 = (a-b)(a^2+ab+b^2)
      Notice:
      1) The sign in the middle of the binomial factor is the same sign as the sign in the original.
      2) The sign on the middle term of the trinomial factor will be the opposite of the sign in the binomial factor.
      3) All other signs are positive.

      Hope this helps.
      (4 votes)
  • leaf red style avatar for user Jack McClelland
    Couldn't you factor the original equation (27x^6+125) as (3x^2+5)^3? Every term is cubed, so why doesn't this work?
    (1 vote)
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  • blobby green style avatar for user Brad Veeder
    how would you factor something with no common variable? For example, (x^3)-3(x^2)+6x-18
    (0 votes)
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Video transcript

Factor 27x to the sixth plus 125. So this is a pretty interesting problem. And frankly, the only way to do this is if you recognize it as a special form. And what I want to do is kind of show you the special form first. And then we can kind of pattern match. So the special form is if I were to take-- and this is really just something you need to know. And I'd argue whether you really need to know this. But actually to do this problem, it's something you just need to know. And that's if you have a squared minus ab plus b squared and you multiply that times a plus b, let's think about what we're going to get. So we're going to take a product right here. We're multiplying. So let's do some algebraic multiplication. So let's multiply. b times b squared is b to the third. b times negative ab is negative ab squared. b times a squared is a squared b. Now let's multiply this top term times a. a times b squared is ab squared. a times negative ab is negative a squared b. And then a times a squared is a to the third. And then we just have to add up all of the terms. We have a positive a squared b and a negative a squared b. So these guys cancel out. We have a negative ab squared and a positive ab squared. These guys cancel out. So all we're left with is an a to the third here and then plus this b to the third. Or another way to think about it-- if someone gives you a to the third plus b to the third, this can be factored into these two expressions. That can be factored into a plus b times a squared minus ab plus b squared. So this is essentially the special form. If you have a sum of cubes, it can be factored out as the sum of the cube roots times this expression right here. And we just showed that it works. So let's see if we have that special form here. Well, 27 is definitely the cube of 3. 3 to the third power is 27. x to the sixth is also the cube of x squared. If you raise x to the sixth to the 1/3 power, you get x squared. So this first term right over here can be rewritten as 3x squared to the third power. And the second term right here-- that's 5 to the third power, so plus 5 to the third power. And this might be a little bit confusing for you. It never hurts to review. Let's multiply 3x squared times 3x squared times 3x squared. That is literally equal to 3 times 3 times 3 times x squared times x squared times x squared. This part right here is 27. x squared times x squared is x to the fourth, times x squared is x to the sixth. Or you could just raise both of these to the third power. 3 to the third is 27. x squared to the third power-- you take an exponent to an exponent, and you're going to take the product of the exponents. So it'll be x to the 2 times 3, or x to the sixth power. So now we know that we have this pattern. So we can just use this. We have the sum of cubes. So just by using this pattern right over here, that means that we can factor it as. This is going to be equal to 3x squared-- that's our a. Let me make it clear. This right here is our a. This right here is our b. So it's going to be a plus b. So it's going to be 3x squared plus b, plus 5, times a squared. Let me do this in a new color. So 3x squared squared-- let's think about that for a second. 3x squared squared-- well, that's going to be 9x to the fourth. So it's going to be times 9x to the fourth minus the product of these two things. So minus the product of 5 and 3x squared, so minus 15x squared. And then finally plus b squared. b is 5. So it's going to be 5 squared, so plus 25. And when I said b is-- this is b, not the whole 5 to the third. And when I say a, just this part is a. And we're done. And I won't explain it in detail in this video. But this right here, if we're thinking about real numbers, we can't actually factor this any more. So we are done factoring this. And remember, this is really just a very, very, very special case of being able to recognize the sum of cubes.