Special Cases - Cubes
Learning Outcomes
- Factor special products
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]
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].
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].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]
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]
Try It
[ohm_question]134492[/ohm_question]- 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]
Contribute!
Licenses & Attributions
CC licensed content, Original
- Revision and Adaptation. Authored by: Lumen Learning. License: CC BY: Attribution.
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.