How Do You Factor a Cubic Equation Like a³c - a³b + b³a - b³c + c³b - c³a?

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The discussion focuses on factoring the cubic equation a³c - a³b + b³a - b³c + c³b - c³a. One participant attempted to factor by grouping but struggled to find a complete solution. They provided a partially simplified form, expressing it as a sum of products involving the variables. Another contributor suggested recognizing the symmetry in the equation and treating it as a polynomial in one variable to collect coefficients. The desired factored form mentioned is (a+b+c)(b-c)(c-a)(a-b), highlighting the challenge of factoring cubic equations with multiple variables.
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Factorising cubic equation

Anyone here know how to factor this equation?

Homework Statement



a^{3}c-a^{3}b+b^{3}a-b^{3}c+c^{3}b-c^{3}a

The Attempt at a Solution



I tried factoring by grouping but ended up getting nowhere.

If anyone can factor this equation, please tell me step by step how you got it.
 
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Cubics are bad enough but this has 3 variables. This is not a complete factorization but I think the most simplified form is
(a^3c- ac^3)+ (b^3a- a^3b)+ (c^3b- b^3c)= ac(a^2- c^2)+ ab(b^2- a^2)+ bc(c^2- b^2)
= ac(a- c)(a+ c)+ ab(b- a)(b+ a)+ bd(c- b)(c+ b)
 
thanks, that will be ok.

but if you could figure how to get this form, please let me know.
(a+b+c)(b-c)(c-a)(a-b)
 
Notice the symmetry in the variables. Then, without loss of generality, treat it as a polynomial in a, and collect coefficients.
 
Yeah, I'd have to agree with Gib Z
 
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