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## Algebra (all content)

### Unit 10: Lesson 27

Advanced polynomial factorization methods# Factoring sum of cubes

CCSS.Math:

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.

## Want to join the conversation?

- 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)- Maybe when you asked that question there wasn't a top level Algebra II group. There is now - just go to https://www.khanacademy.org/math/algebra2(9 votes)

- what if you have a problem that looks like x^3+8 ?(7 votes)
- Then, using the method in this video, that would be (x+2)(x^2-2x+4).(16 votes)

- 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)- I believe 27x^6 factored is (3x^2)^3. The coefficient, 27, is a perfect cube of 3 and x^6 is a cube of x^2.(8 votes)

- Why is there a - in front of the parentheses of (3x^2+5)

I believe thats a mistake(4 votes)- Nope it's an equals sign. At3:24he draws two lines on top of each other, it's pretty messy.(6 votes)

- Why is there a negative sign at3:26? There shouldn't be a negative sign there.(4 votes)
- 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)- 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)

- 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)- 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)

- Any tips on remembering where exactly the minus sign goes in the sum of cubes formula? Same with difference of cubes..(1 vote)
- 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)

- 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)
- Exponents don't distribute over addition. Try using the distributive property to multiply together three (3x^2 + 5)s and see that you don't get the desired result.(3 votes)

- how would you factor something with no common variable? For example, (x^3)-3(x^2)+6x-18(0 votes)
- You would factor this by grouping

1 - group in two sets

(x^3-3x^2)+(6x-18)

2 - Pull out the common factor of each set

x^2(x-3)+6(x-3)

3 - Then pull out the common expression

(x-3)(x^2+6)

Note that this only works if the two expressions from step 2 match(5 votes)

## 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.