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Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

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BuyFindarrow_forward

Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

For Problems 51 68 , use the sum-of-two-cubes or the difference-of-two-cubes pattern to factor each of the following. (Objective 2)

x 6 y 6 .

To determine

To factor:

The given problem by the use of the sum-of-cubes pattern or the difference-of-cubes pattern.

Explanation

Approach:

i) The Difference of Two Squares pattern is a2b2=(a+b)(ab).

ii) The Sum of Two Cubes pattern is a3+b3=(a+b)(a2ab+b2).

iii) The Difference of Two Squares pattern is a3b3=(ab)(a2+ab+b2).

iv) Multiplication is commutative, so the order of writing the factors is not important. For example, (ab)(a2+ab+b2) can also be written as (a2+ab+b2)(ab).

The following steps have been followed to find out the factors:

Step-1: Convert the polynomial in the form of 2 squares with subtraction and compare with the difference of two squares pattern.

Step-2: Replace the value of a and b in the difference of two squares pattern by the binomial values available from the step-1.

Step-3: If the equation from step-2 has the sum of two cubes pattern, then compare and replace the value of a and b.

Step-4: For the polynomial in the form of the difference of two cubes pattern available in step-2, compare and replace the value of a and b.

Step-5: Write the complete factors of the given polynomial and simplify.

Calculation:

The given polynomial is x6y6.

It can be written as,

x6y6=(x3)2(y3)2

Because, (x3)2=x3x3=x3+3=x6 and (y3)2=y3y3=y3+3=y6.

The above equation represents the difference of two squares pattern. That is, a2b2=(a+b)(ab).

Replace the value of the difference of two squares pattern by the above polynomial elements.

That is, a=x3 and b=y3.

Therefore the factors of the polynomial are as follows:

(x3)2(y3)2=(x3+y3)(x3y3)

In the above equation, the polynomial (x3+y3) has 2 cubes and they are added. So, it represents the sum of two cubes pattern. That is, a3+b3=(a+b)(a2ab+b2).

Replace the value of the difference of two cubes pattern as, a=x and b=y

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P-48CRSect-3.CR P-49CRSect-3.CR P-50CRSect-3.CR P-51CRSect-3.CR P-52CRSect-3.CR P-53CRSect-3.CR P-54CRSect-3.CR P-55CRSect-3.CR P-56CRSect-3.CR P-57CRSect-3.CR P-58CRSect-3.CR P-59CRSect-3.CR P-60CRSect-3.CR P-61CRSect-3.CR P-62CRSect-3.CR P-63CRSect-3.CR P-64CRSect-3.CR P-65CRSect-3.CR P-66CRSect-3.CR P-67CRSect-3.CR P-68CRSect-3.CR P-69CRSect-3.CR P-70CRSect-3.CR P-71CRSect-3.CR P-72CRSect-3.CR P-73CRSect-3.CR P-74CRSect-3.CR P-75CRSect-3.CR P-76CRSect-3.CR P-77CRSect-3.CR P-78CRSect-3.CR P-79CRSect-3.CR P-80CRSect-3.CR P-81CRSect-3.CR P-82CRSect-3.CR P-83CRSect-3.CR P-84CRSect-3.CR P-85CRSect-3.CR P-86CRSect-3.CR P-87CRSect-3.CR P-88CRSect-3.CR P-89CRSect-3.CR P-90CRSect-3.CR P-91CRSect-3.CR P-92CRSect-3.CR P-93CRSect-3.CR P-94CRSect-3.CR P-95CRSect-3.CR P-96CRSect-3.CR P-97CRSect-3.CR P-98CRSect-3.CR P-99CRSect-3.CR P-100CRSect-3.CR P-101CRSect-3.CR P-102CRSect-3.CR P-103CRSect-3.CR P-104CRSect-3.CR P-105CRSect-3.CR P-106CRSect-3.CR P-107CRSect-3.CR P-108CRSect-3.CR P-109CRSect-3.CR P-110CRSect-3.CR P-111CRSect-3.CR P-112CRSect-3.CR P-113CRSect-3.CR P-114CRSect-3.CT P-1CTSect-3.CT P-2CTSect-3.CT P-3CTSect-3.CT P-4CTSect-3.CT P-5CTSect-3.CT P-6CTSect-3.CT P-7CTSect-3.CT P-8CTSect-3.CT P-9CTSect-3.CT P-10CTSect-3.CT P-11CTSect-3.CT P-12CTSect-3.CT P-13CTSect-3.CT P-14CTSect-3.CT P-15CTSect-3.CT P-16CTSect-3.CT P-17CTSect-3.CT P-18CTSect-3.CT P-19CTSect-3.CT P-20CTSect-3.CT P-21CTSect-3.CT P-22CTSect-3.CT P-23CTSect-3.CT P-24CTSect-3.CT P-25CT

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