What is the factored form of 8x^24 – 27y^6

What is the factored form of 8x^24 - 27y^6

Answers

(2x8 - 3y2) • (4x16 + 6x8y2 + 9y4)

Step-by-step explanation:

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Step-by-step explanation:

 The required factored form of the given expression is (2x^8-3y^2)(4x^{16}+6x^8y^2+9y^4).

Step-by-step explanation:  We are given to find the factored form of the following expression :

E=8x^{24}-27y^6.

We will be using the following factorization formula :

a^3-b^3=(a-b)(a^2+ab+b^2).

So, the factorization of the given expression is as follows :

E\\=8x^{24}-27y^6\\=(2x^8)^3-(3y^2)^3\\=(2x^8-3y^2)((2x^8)^2+2x^8times3y^2+(3y^2)^2)\\=(2x^8-3y^2)(4x^{16}+6x^8y^2+9y^4).

Thus, the required factored form of the given expression is (2x^8-3y^2)(4x^{16}+6x^8y^2+9y^4).


 8x24-27y6 Final result : (2x8 - 3y2) • (4x16 + 6x8y2 + 9y4) Step by step solution : Step  1  :Skip Ad
Equation at the end of step  1  : (8 • (x24)) - 33y6 Step  2  :Equation at the end of step  2  : 23x24 - 33y6 Step  3  :Trying to factor as a Difference of Squares :

 3.1      Factoring:  8x24-27y6 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 - AB + AB - B2 =
         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  8  is not a square !!

Ruling : Binomial can not be factored as the
difference of two perfect squares

Trying to factor as a Difference of Cubes:

 3.2      Factoring:  8x24-27y6 

Theory : A difference of two perfect cubes,  a3 - b3 can be factored into
              (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =
            a3+a2b+ab2-ba2-b2a-b3 =
            a3+(a2b-ba2)+(ab2-b2a)-b3 =
            a3+0+0+b3 =
            a3+b3

Check :  8  is the cube of  2 

Check :  27  is the cube of   3 
Check :  x24 is the cube of   x8

Check :  y6 is the cube of   y2

Factorization is :
             (2x8 - 3y2)  •  (4x16 + 6x8y2 + 9y4) 

Trying to factor as a Difference of Squares :

 3.3      Factoring:  2x8 - 3y2 

Check :  2  is not a square !!

Ruling : Binomial can not be factored as the
difference of two perfect squares

Trying to factor a multi variable polynomial :

 3.4    Factoring    4x16 + 6x8y2 + 9y4 

Try to factor this multi-variable trinomial using trial and error 

 Factorization fails

Final result : (2x8 - 3y2) • (4x16 + 6x8y2 + 9y4)

Factoring:  8x^{24}-27y^6  


Theory : A difference of two perfect cubes,  a^3-b^3 can be factored into

             (a-b)*(a^2+ab+b^2)


Proof :  (a-b)*(a^2+ab+b^2)=

           a^3+a^2b+ab^2-ba^2-b^2a-b^3=

           a^3+(a^2b-ba^2)+(ab^2-b^2a)-b^3=

           a^3+0+0+b^3=

           a^3+b^3


Check :  8  is the cube of  2  


Check :  27  is the cube of   3  

Check :   x^{24} is the cube of    x^8


Check :  y^6 is the cube of    y^2


Factorization is :

            (2x^8 - 3y^2)*(4x^{16} + 6x^8y^2 + 9y^4)

The answer is x^6(2X^6-3)(4X^12+6X^6+9)
The answer will be (2x8 - 3y2) • (4x16 + 6x8y2 + 9y4)
The factored form of that expression is,
(2x^8-3y^2)(4x^16+6x^8y^2+9y^4).

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