SOLUTION: Factor ab^3 - a^3b + bc^3 - b^3c + ca^3 - c^3a. How would I go about doing this problem? Thanks!

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Question 1175216: Factor
ab^3 - a^3b + bc^3 - b^3c + ca^3 - c^3a. How would I go about doing this problem? Thanks!

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


This apparently requires a factorization technique that I am not familiar with.

I went to wolframalpha.com and entered "factor ab^3 - a^3b + bc^3 - b^3c + ca^3 - c^3a" and it did it:



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If you're not familiar with wolframalpha.com, you should be. It's amazing -- it will do almost anything....


Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

I'd say 


    If you are an average student and if you don't know the solution in advance, 

    you will be not able to complete this assignment without help from outside

         (without help from some outer source as Wikipedia; or some advanced textbook; or some web-site).



For more advanced Algebra student, the way of thinking can be THIS:


    a)  you have a polynomial of the uniform degree 4;

        hence, you may expect having the product of four homogeneous linear binomials;


    b)  if you change "a" for "b" everywhere in the left side, left side 
        will change its sign.


        If you change "a" for "c" everywhere in the left side, left side 
        will change its sign.


        If you change "b" for "c" everywhere in the left side, left side 
        will change its sign.


        Hence, the advanced student can foresee that the factored form should contain the factors (a-b), (a-c) and (b-c).


        Thus such student just sees the product  (a-b)*(a-c)*(b-c)  through the fog at the horizon.


     c)  And then the student must make a super-human effort to guess that the last multiplier is (a+b+c).


To get a success, such a student should be trained analyzing symmetric (or antisymmetric)
homogeneous functions of several variables.

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For more information, see this link

https://math.stackexchange.com/questions/1379045/how-to-factor-intricate-polynomial-ab3-a3b-a3c-ac3-bc3-b3c



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