Solution - Factoring multivariable polynomials

Step  1  :

Trying to factor as a Difference of Cubes:

 1.1      Factoring:  y³-z³ 

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 :  y³ is the cube of   y¹

Check :  z³ is the cube of   z¹

Factorization is :
             (y - z)  •  (y² + yz + z²) 

Trying to factor a multi variable polynomial :

 1.2    Factoring    y² + yz + z² 

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

 Factorization fails

Equation at the end of step  1  :

  (((x•(y-z)•(y2+yz+z2)•y)•(z3-x3))•z)•(x3-y3)

Step  2  :

Equation at the end of step  2  :

  ((xy•(y-z)•(y2+yz+z2)•(z3-x3))•z)•(x3-y3)

Step  3  :

Trying to factor as a Difference of Cubes:

 3.1      Factoring:  z³-x³ 

Check :  z³ is the cube of   z¹

Check :  x³ is the cube of   x¹

Factorization is :
             (z - x)  •  (z² + xz + x²) 

Trying to factor a multi variable polynomial :

 3.2    Factoring    z² + xz + x² 

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

 Factorization fails

Equation at the end of step  3  :

  (xy•(y-z)•(y2+yz+z2)•(z-x)•(x2+xz+z2)•z)•(x3-y3)

Step  4  :

Equation at the end of step  4  :

  xyz•(y-z)•(y2+yz+z2)•(z-x)•(x2+xz+z2)•(x3-y3)

Step  5  :

Trying to factor as a Difference of Cubes:

 5.1      Factoring:  x³-y³ 

Check :  x³ is the cube of   x¹

Check :  y³ is the cube of   y¹

Factorization is :
             (x - y)  •  (x² + xy + y²) 

Trying to factor a multi variable polynomial :

 5.2    Factoring    x² + xy + y² 

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

 Factorization fails

Final result :

  xyz•(y-z)•(y2+yz+z2)•(z-x)•(x2+xz+z2)•(x-y)•(x2+xy+y2)