Is it possible to factor the polynomial Factor the polynomia

Is it possible to factor the polynomial?

Factor the polynomial. 64a^3 - 27b^6

Solution

given

64a^3 - 27b^6

Step  1  :

Equation at the end of step  1  :

Step 2 :

Equation at the end of step  2  :

Step  3  :

Trying to factor as a Difference of Squares :

3.1      Factoring:  64a3-27b6

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 :  64  is the square of  8
Check : 27 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:  64a3-27b6

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 :  64  is the cube of  4

Check :  27  is the cube of   3
Check : a3 is the cube of   a1

Check : b6 is the cube of   b2

Factorization is :
             (4a - 3b2)  •  (16a2 + 12ab2 + 9b4)

Trying to factor as a Difference of Squares :

3.3      Factoring:  4a - 3b2

Check :  4  is the square of  2
Check : 3 is not a square !!

Factoring    16a2 + 12ab2 + 9b4 not possible

final result:

Is it possible to factor the polynomial? Factor the polynomial. 64a^3 - 27b^6Solutiongiven 64a^3 - 27b^6 Step 1 : Equation at the end of step 1 : Step 2 : Equat

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