In many ways, factoring is about patterns—if you identify the trends that number make once they room multiplied together, you have the right to use those trends to separate these numbers right into their separation, personal, instance factors.

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Some exciting patterns arise once you are working v cubed amounts within polynomials. Special, there room two an ext special instances to consider: a3 + b3 and also a3 – b3.

Let’s take it a watch at exactly how to element sums and also differences that cubes.


The hatchet “cubed” is provided to describe a number increased to the 3rd power. In geometry, a cube is a six-sided form with equal width, length, and height; since all these actions are equal, the volume of a cube with width x have the right to be stood for by x3. (Notice the exponent!)

Cubed numbers get big very quickly. 13 = 1, 23 = 8, 33 = 27, 43 = 64, and 53 = 125.

Before looking in ~ factoring a sum of 2 cubes, stop look in ~ the feasible factors.

It transforms out that a3 + b3 deserve to actually it is in factored together (a + b)(a2 – ab + b2). Let’s inspect these components by multiplying.


does (a + b)(a2 ab + b2) = a3 + b3?

(a)(a2 – abdominal + b2) + (b)(a2 – ab +b2)

Apply the distributive property.

(a3 – a2b + ab2) + (b)(a2 - abdominal + b2)

Multiply by a.

(a3 – a2b + ab2) + (a2b – ab2 + b3)

Multiply through b.

a3 – a2b + a2b + ab2 – ab2 + b3

Rearrange terms in stimulate to combine the choose terms.

a3 + b3

Simplify


Did you view that? four of the terms cancelled out, leaving us with the (seemingly) basic binomial a3 + b3. So, the components are correct.

You have the right to use this sample to aspect binomials in the type a3 + b3, otherwise recognized as “the amount of cubes.”

The sum of Cubes

A binomial in the type a3 + b3 can be factored together (a + b)(a2 – abdominal muscle + b2).

Examples:

The factored kind of x3 + 64 is (x + 4)(x2 – 4x + 16).

The factored kind of 8x3 + y3 is (2x + y)(4x2 – 2xy + y2).


Example

Problem

Factor x3 + 8y3.

x3 + 8y3

Identify the this binomial fits the sum of cubes pattern: a3 + b3.

a = x, and b = 2y (since 2y • 2y • 2y = 8y3).

(x + 2y)(x2 – x(2y) + (2y)2)

Factor the binomial as

(a + b)(a2 – abdominal + b2), substituting a = x and also b = 2y right into the expression.

(x + 2y)(x2 – x(2y) + 4y2)

Square (2y)2 = 4y2.

Answer

(x + 2y)(x2 – 2xy + 4y2)

Multiply −x(2y) = −2xy (writing the coefficient first.


And that’s it. The binomial x3 + 8y3 deserve to be factored as (x + 2y)(x2 – 2xy + 4y2)! Let’s shot another one.

You should always look because that a common factor prior to you follow any of the fads for factoring.


Example

Problem

Factor 16m3 + 54n3.

16m3 + 54n3

Factor the end the usual factor 2.

2(8m3 + 27n3)

8m3 and 27n3 are cubes, so friend can factor 8m3 + 27n3 together the amount of two cubes: a = 2m, and also b = 3n.

2(2m + 3n)<(2m)2 – (2m)(3n) + (3n)2>

Factor the binomial 8m3 + 27n3 substituting a = 2m and b = 3n into the expression (a + b)(a2 – abdominal muscle + b2).

2(2m + 3n)<4m2 – (2m)(3n) + 9n2>

Square: (2m)2 = 4m2 and (3n)2 = 9n2.

Answer

2(2m + 3n)(4m2 – 6mn + 9n2)

Multiply −(2m)(3n) = −6mn.


Factor 125x3 + 64.

A) (5x + 64)(25x2 – 125x + 16)

B) (5x + 4)(25x2 – 20x + 16)

C) (x + 4)(x2 – 2x + 16)

D) (5x + 4)(25x2 + 20x – 64)


Show/Hide Answer

A) (5x + 64)(25x2 – 125x + 16)

Incorrect. Inspect your values for a and b here. B3 = 64, therefore what is b? The correct answer is (5x + 4)(25x2 – 20x + 16).

B) (5x + 4)(25x2 – 20x + 16)

Correct. 5x is the cube source of 125x3, and also 4 is the cube root of 64. Substituting these worths for a and also b, you uncover (5x + 4)(25x2 – 20x + 16).

C) (x + 4)(x2 – 2x + 16)

Incorrect. Examine your values for a and also b here. A3 = 125x3, therefore what is a? The correct answer is (5x + 4)(25x2 – 20x + 16).

D) (5x + 4)(25x2 + 20x – 64)

Incorrect. Check the mathematics signs; the b2 ax is positive, not negative, as soon as factoring a amount of cubes. The correct answer is (5x + 4)(25x2 – 20x + 16).

Difference the Cubes


Having seen just how binomials in the kind a3 + b3 can be factored, it need to not come together a surprise that binomials in the kind a3 – b3 can be factored in a similar way.

The difference of Cubes

A binomial in the form a3 – b3 can be factored together (a – b)(a2 + abdominal + b2).

Examples:

The factored kind of x3 – 64 is (x – 4)(x2 + 4x + 16).

The factored kind of 27x3 – 8y3 is (3x – 2y)(9x2 + 6xy + 4y2).

Notice that the basic construction the the administrate is the exact same as the is for the amount of cubes; the difference is in the + and also – signs. Take a minute to compare the factored kind of a3 + b3 through the factored kind of a3 – b3.


Factored type of a3 + b3:

(a + b)(a2 – ab + b2)

Factored form of a3 – b3:

(a – b)(a2 + abdominal + b2)


This deserve to be tricky come remember due to the fact that of the different signs—the factored kind of a3 + b3 contains a negative, and also the factored type of a3 – b3 contains a positive! Some human being remember the different forms like this:

“Remember one succession of variables: a3 b3 = (a b)(a2 ab b2). There space 4 absent signs. Whatever the very first sign is, it is additionally the 2nd sign. The 3rd sign is the opposite, and the 4th sign is always +.”

Try this for yourself. If the an initial sign is +, together in a3 + b3, follow to this strategy just how do you fill in the rest: (a b)(a2 abdominal muscle b2)? walk this an approach help you remember the factored type of a3 + b3 and a3 – b3?

Let’s walk ahead and look in ~ a pair of examples. Mental to variable out all common factors first.


Example

Problem

Factor 8x3 – 1,000.

8(x3 – 125)

Factor the end 8.

8(x3 – 125)

Identify the the binomial fits the pattern a3 - b3: a = x, and also b = 5 (since 53 = 125).

8(x - 5)

Factor x3 – 125 as (a – b)(a2 + abdominal muscle + b2), substituting a = x and also b = 5 right into the expression.

8(x – 5)(x2 + 5x + 25)

Square the an initial and last terms, and also rewrite (x)(5) as 5x.

Answer

8(x – 5)(x2 + 5x + 25)


Let’s see what happens if friend don’t factor out the typical factor first. In this example, it have the right to still be factored together the difference of 2 cubes. However, the factored kind still has typical factors, which have to be factored out.


Example

Problem

Factor 8x3 – 1,000.

8x3 – 1,000

Identify the this binomial fits the pattern a3 - b3: a = 2x, and b = 10 (since 103 = 1,000).

(2x – 10)<(2x)2 + 2x(10) + 102>

Factor as (a – b)(a2 + abdominal muscle + b2), substituting a = 2x and b = 10 right into the expression.

(2x – 10)(4x2 + 20x + 100)

Square and also multiply: (2x)2 = 4x2,

(2x)(10) = 20x, and 102 = 100.

2(x – 5)(4)(x2 + 5x + 25)

Factor out remaining common factors in every factor. Variable out 2 indigenous the an initial factor, aspect out 4 native the second factor.

(2 • 4)(x – 5)(x2 + 5x + 25)

Multiply the numerical factors.

Answer

8(x – 5)(x2 + 5x + 25)


As you have the right to see, this last example still worked, but required a pair of extra steps. It is constantly a an excellent idea to aspect out all typical factors first. In part cases, the just efficient method to factor the binomial is to element out the common factors first.

Here is one an ext example. Note that r9 = (r3)3 and also that 8s6 = (2s2)3.


Example

Problem

Factor r9 – 8s6.

r9 – 8s6

Identify this binomial together the difference of 2 cubes. As shown above, that is. Utilizing the legislations of exponents, rewrite r9 together (r3)3.

(r3)3 – (2s2)3

Rewrite r9 as (r3)3 and rewrite 8s6 together (2s2)3.

Now the binomial is written in regards to cubed quantities. Reasoning of a3 – b3,

a = r3 and also b = 2s2.

(r3 – 2s2)<(r3)2 + (r3)(2s2) + (2s2)2>

Factor the binomial as

 (a – b)(a2 + ab + b2), substituting a = r3 and b = 2s2 right into the expression.

(r3 – 2s2)(r6 + 2 r3s2+ 4s4)

Multiply and square the terms.

Answer

(r3 – 2s2)(r6 + 2r3s2 + 4s4)


Using the difference of cubes, determine the product the 3(x – 3y)(x2 + 3xy + 9y2).

A) x3 – y3

B) 3x – 81y

C) 3x3 + 81y3

D) 3x3 – 81y3


Show/Hide Answer

A) x3 – y3

Incorrect. If this to be true, the expression shown over would it is in (x – y)(x2 + xy + y2). The correct answer is 3x3 – 81y3.

B) 3x – 81y

Incorrect. No of the terms in this binomial is a cubed number! The exactly answer is

3x3 – 81y3.

C) 3x3 + 81y3

Incorrect. Examine your signs. If this expression drops into the distinction of cubes category, the symbol in between 3x3 and 81y3 have to be –. The correct answer is 3x3 – 81y3.

See more: -4 To The 4Th Power ? Solution: Find The Value Of 4 To The 4Th Power

D) 3x3 – 81y3

Correct. Recognizing that this expression is in the kind (a – b)(a2 + ab + b2), you uncover a = x and b = 3y. This way that the resulting a3 – b3 monomial is x3 – 27y3. It additionally needs come be multiply by the coefficient 3: 3x3 – 81y3.

Summary


You conference some interesting patterns once factoring. Two special cases—the amount of cubes and the difference of cubes—can aid you factor some binomials that have a level of 3 (or higher, in part cases). The special situations are:

A binomial in the form a3 + b3 deserve to be factored as (a + b)(a2 – abdominal + b2) A binomial in the form a3 – b3 can be factored as (a – b)(a2 + ab + b2)