If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra (all content)>Unit 10

Lesson 27: Advanced polynomial factorization methods

# Factoring difference of cubes

Sal factors 40c^3-5d^3 as 5(2x-d)(4c^2+2cd+d^2) using a special product form for a difference of cubes. Created by Sal Khan.

## Want to join the conversation?

• How to solve the equation:x^3-6x^2+9x=4? The x-values when y is 4 is 1 and 4. I don't know how to get that.
• how to prove the accuracy of the formula for difference of cube?
(1 vote)
• So we have a^3 - b^3 = (a-b)(a^2 + ab + b^2)
If you simply multiply out these brackets you will get a^3 - b^3, so it is accurate.
• what is this 5(2c-d)(4c^2+2cd+d^2) for ? and why did we factor it out?
• We factor it because it makes it simpler to find solutions. If we are dealing with the equation 40c^3 - 5d^3 = 0, for example, it is not easy to find the solutions of a cubic equation, so we factorise it into a linear equation and a quadratic, both of which we can find solutions for very easily.
• How would you factor 5x^3-40 ? and 4x^10-y^6?
• Okay, so I understand that the rule is, if you get something like
a^3 + b^3,
the factored version is gonna have a '-' between the first and second terms in the trinomial, and if you get
a^3 - b^3,
you'll get a '+' between those terms.
Either way, there's always a '+' between the second and third terms.

What happens though, if you get a problem where both cubed terms are negative?
Does THAT give you a '-' between the second and third terms in the factored form?

(I hope I worded this question okay.)
• If both terms are negative like: -a^3 - b^3
1) Factor out a common factor of -1: -1 (a^3+b^3)
2) then, factor the sum of cubes: -1 (a+b) (a^2-ab+b^2)
Hope this helps.
• I am confused. I know that the video says that "d" is the b value, but couldn't you use "-d" as your b value in the sum of cubes? Or when you do the difference of cubes is it supposed to be "-d" Help
• Either way would be fine. Difference of cubes is not really a separate rule, but it can seem a bit confusing since difference of squares and sum of squares are different rules. But your intuition is correct. You could write the sample polynomial as 5((2c)^3 + (-d)^3)
• i have sort of a similar problem that asks me to factor COMPLETELY
x^6 - y^6 how to would i completely factor this after using the differences of cubes
(1 vote)
• Actually, you can go one more step, I believe. You can then use the difference of cubes for (x^3-y^3), and can use sum of cubes on (x^3+y^3). Now that I think about it, it may not be getting any further...

You should get (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)
• Why is factoring cubes so difficult? :(
(1 vote)
• You need to learn the pattern. It is confusing initially until you understand how to use the pattern and have it memorized.
• Could you fist simplify it to -5((-2c)^3+d^3) and factor it from there or is one way better?
(1 vote)
• There is no need to factor out a -5. Usually you would factor out a negative when the lead coefficient is negative. That is not the case here.

You have changed the difference of 2 cubes into a sum of 2 cubes. Both are factorable, but use slightly different patterns.