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Solution - Factoring binomials using the difference of squares

(x3y+1)(x3y1)
(x^3y+1)*(x^3y-1)

Step by Step Solution

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "y2"   was replaced by   "y^2".  1 more similar replacement(s).

Step  1  :

Trying to factor as a Difference of Squares :

 1.1      Factoring:  x6y2-1 

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 : 1 is the square of 1
Check :  x6  is the square of  x3 

Check :  y2  is the square of  y1 

Factorization is :       (x3y + 1)  •  (x3y - 1) 

Trying to factor as a Sum of Cubes :

 1.2      Factoring:  x3y + 1 

Theory : A sum 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 :  1  is the cube of   1 
Check :  x3 is the cube of   x1

Check :  y 1 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes

Trying to factor as a Difference of Cubes:

 1.3      Factoring: x3y - 1

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 :  1  is the cube of   1 
Check :  x3 is the cube of   x1

Check :  y 1 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

  (x3y + 1) • (x3y - 1)

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