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Factor ThisSubmitted by Maria Elena Alvarez, 04/04/1997. Original answer by Valerio De Angelis; this article by Doug Dillon and Allen Stenger.Factor the expression A3(B-C) + B3(C-A) + C3(A-B). Where does one even start?? Hint 1Note that the expression is zero if B = C, or if A = C, or if A = B. This means that B - C, C - A, A - B must all be factors. If you factor these out, what is left over? Hint 2There are a couple of ways to find the remaining factor. The brute-force method is to multiply out the original expression, use polynomial long division to take out the known factors, then see what is left over. But there is another way you can reason out the remaining factor. Look at the symmetries in the original expression and in the known factors; what do they tell us about the remaining factors? Hint 3The original expression has a circular symmetry in A, B, C; that is, if you substitute A for B, B for C, and C for A, you get the same expression over again. If you make this substitution in the known factors (B-C)(C-A)(A-B) you also get the same expression over again. Therefore what? The Rest of the SolutionThe remaining factor is the quotient of two expressions that have the circular symmetry property, so it must also have the circular symmetry property. Because it is linear, the only possibility is that it is a constant factor times (A + B + C), that is, k(A + B + C), and the whole factorization is A3(B-C) + B3(C-A) + C3(A-B) = k(B-C)(C-A)(A-B)(A+B+C). We can then find k by examining (for example) the coefficient of A3B, which on the left is 1 and on the right is -k, so k = -1, and the complete factorization is A3(B-C) + B3(C-A) + C3(A-B) = -(B-C)(C-A)(A-B)(A+B+C). Another Approach: Where Did This Problem Come From?This problem can be summarized as "consider the following horrible-looking expression and figure out something about it." There are thousands (maybe millions) of math problems and puzzles that have this form. The author of the problem knows where the expression came from, but you don't. Often if you can guess where it came from, you can work the problem in a simple way. We've already noted that there is a circular symmetry in the variables, and that the expression is 0 if any two variables are equal. It's also true that there is a transposition anti-symmetry: if you swap any two variables, you negate the value of the expression. These properties should make you start thinking about determinants, because they have many of these same properties, and you should think about whether the expression is the expansion of some determinant. A little experimentation shows that it is equal to the following determinant, as you can see by expanding by minors along the top row:
We don't know if that's where the expression came from, but it is interesting, especially since we see that we can evaluate the determinant in another way to eliminating the 1s along the bottom row:
Symmetries and InvarianceMany mathematical and physical problems have natural symmetries, and you can often use these to help solve the problem. One area that makes especially heavy use of symmetries is the Theory of Equations and Galois Theory. The key observation is that the coefficients of a polynomial are symmetric functions of its roots; for example, if the roots of x2 - ax + b = 0 are r1, r2, so that x2 - ax + b = (x - r1)(x - r2), then a = r1 + r2 and b = r1 r2 are symmetric functions of the roots. Galois Theory uses permutation groups to study the symmetries of functions of the roots. The determinant we used in this solution is similar to the Vandermonde determinant that appears in several areas of mathematics, including the Theory of Equations. The Vandermonde determinant in three variables is
You can work out an expression for the three-variable Vandermonde determinant in the same way as before, and you will get (B-A)(C-A)(C-B). The square of the Vandermonde determinant in any number of variables is a symmetric function of its variables; if the variables are the roots of a polynomial, then the square can be written in terms of the coefficients of the polynomial, and it is called the discriminant of the polynomial ("discriminant" because it tells you whether the polynomial has any repeated roots). You can read more about symmetric functions, the Theory of Equations, Galois Theory, and the Vandermonde determinant in most books on Abstract Algebra, for example in Clark's book cited below. References
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