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Study Guides > ALGEBRA / TRIG I

Special Cases - Cubes

Learning Outcomes

  • Factor special products
Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}-b^{3}[/latex]. Let us take a look at how to factor sums and differences of cubes.

Sum of Cubes

The term “cubed” is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width [latex]x[/latex] can be represented by [latex]x^{3}[/latex]. (Notice the exponent!) Cubed numbers get large very quickly: [latex]1^{3}=1[/latex], [latex]2^{3}=8[/latex], [latex]3^{3}=27[/latex], [latex]4^{3}=64[/latex], and [latex]5^{3}=125[/latex] Before looking at factoring a sum of two cubes, let us look at the possible factors. It turns out that [latex]a^{3}+b^{3}[/latex] can actually be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]. Check these factors by multiplying.

Example

Does [latex](a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}[/latex]?

Answer: Apply the distributive property.

[latex]\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)[/latex]

Multiply by a.

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)[/latex]

Multiply by b.

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)[/latex]

Rearrange terms in order to combine the like terms.

[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[/latex]

Simplify.

[latex]a^{3}+b^{3}[/latex]

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[/latex]. So, the factors are correct. You can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[/latex], otherwise known as “the sum of cubes.”

The Sum of Cubes

A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex].

Examples

The factored form of [latex]x^{3}+64[/latex] is [latex]\left(x+4\right)\left(x^{2}–4x+16\right)[/latex]. The factored form of [latex]8x^{3}+y^{3}[/latex] is [latex]\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)[/latex].

Example

Factor [latex]x^{3}+8y^{3}[/latex].

Answer: Identify that this binomial fits the sum of cubes pattern [latex]a^{3}+b^{3}[/latex]. [latex]a=x[/latex], and [latex]b=2y[/latex] (since [latex]2y\cdot2y\cdot2y=8y^{3}[/latex]).

[latex]x^{3}+8y^{3}[/latex]

Factor the binomial as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=2y[/latex] into the expression.

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)[/latex]

Square [latex](2y)^{2}=4y^{2}[/latex].

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)[/latex]

Multiply [latex]−x\left(2y\right)=−2xy[/latex] (writing the coefficient first).

The factored form is [latex]\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)[/latex].

Now try another one. You should always look for a common factor before you follow any of the patterns for factoring.

Example

Factor [latex]16m^{3}+54n^{3}[/latex].

Answer: Factor out the common factor [latex]2[/latex].

[latex]16m^{3}+54n^{3}[/latex]

[latex]2\left(8m^{3}+27n^{3}\right)[/latex]

[latex]8m^{3}[/latex] and [latex]27n^{3}[/latex] are cubes, so you can factor [latex]8m^{3}+27n^{3}[/latex] as the sum of two cubes: [latex]a=2m[/latex] and [latex]b=3n[/latex]. Factor the binomial [latex]8m^{3}+27n^{3}[/latex] substituting [latex]a=2m[/latex] and [latex]b=3n[/latex] into the expression [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex].

[latex]2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right][/latex]

Square: [latex](2m)^{2}=4m^{2}[/latex] and [latex](3n)^{2}=9n^{2}[/latex].

[latex]2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right][/latex]

Multiply [latex]-\left(2m\right)\left(3n\right)=-6mn[/latex].

The factored form is [latex]2\left(2m+3n\right)\left(4m^{2}-6mn+9n^{2}\right)[/latex].

Try It

[ohm_question]1952[/ohm_question]

Difference of Cubes

Having seen how binomials in the form [latex]a^{3}+b^{3}[/latex] can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[/latex] can be factored in a similar way.

The Difference of Cubes

A binomial in the form [latex]a^{3}–b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex].

Examples

The factored form of [latex]x^{3}–64[/latex] is [latex]\left(x–4\right)\left(x^{2}+4x+16\right)[/latex]. The factored form of [latex]27x^{3}–8y^{3}[/latex] is [latex]\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)[/latex].
Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[/latex] and [latex]–[/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[/latex] with the factored form of [latex]a^{3}-b^{3}[/latex]. Factored form of [latex]a^{3}+b^{3}[/latex]: [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex] Factored form of [latex]a^{3}-b^{3}[/latex]: [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex] This can be tricky to remember because of the different signs. The factored form of [latex]a^{3}+b^{3}[/latex] contains a negative, and the factored form of [latex]a^{3}-b^{3}[/latex] contains a positive! Some people remember the different forms like this: “Remember one sequence of variables: [latex]a^{3}b^{3}=\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]. There are [latex]4[/latex] missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[/latex].” Try this for yourself. If the first sign is [latex]+[/latex], as in [latex]a^{3}+b^{3}[/latex], according to this strategy, how do you fill in the rest: [latex]\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]? Does this method help you remember the factored form of [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}–b^{3}[/latex]? Let us go ahead and look at a couple of examples. Remember to factor out all common factors first.

Example

Factor [latex]8x^{3}–1,000[/latex].

Answer: Factor out [latex]8[/latex].

[latex]8(x^{3}–125)[/latex]

Identify that the binomial fits the pattern [latex]a^{3}-b^{3}:a=x[/latex], and [latex]b=5[/latex] (since [latex]5^{3}=125[/latex]). Factor [latex]x^{3}–125[/latex] as [latex]\left(a–b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=5[/latex] into the expression.

[latex]8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right][/latex]

Square the first and last terms, and rewrite [latex]\left(x\right)\left(5\right)[/latex] as [latex]5x[/latex].

[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]

Here is one more example. Note that [latex]r^{9}=\left(r^{3}\right)^{3}[/latex] and that [latex]8s^{6}=\left(2s^{2}\right)^{3}[/latex].

Example

Factor [latex]r^{9}-8s^{6}[/latex].

Answer: Identify this binomial as the difference of two cubes. As shown above, it is. Rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex] and rewrite [latex]8s^{6}[/latex] as [latex]\left(2s^{2}\right)^{3}[/latex].

[latex]\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}[/latex]

Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[/latex], [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex]. Factor the binomial as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex] into the expression.

[latex]\left(r^{3}-2s^{2}\right)\left[\left(r^{3}\right)^{2}+\left(r^{3}\right)\left(2s^{2}\right)+\left(2s^{2}\right)^{2}\right][/latex]

Multiply and square the terms.

[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]

In the following two video examples, we show more binomials that can be factored as a sum or difference of cubes. https://youtu.be/tFSEpOB262M https://youtu.be/J_0ctMrl5_0

Try It

[ohm_question]134492[/ohm_question]
You encounter some interesting patterns when factoring. Two special cases—the sum of cubes and the difference of cubes—can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:
  • A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]
  • A binomial in the form [latex]a^{3}-b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex]
Always remember to factor out any common factors first.

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Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • Ex 3: Factor a Sum or Difference of Cubes. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.