factorise a^5(b-c)+b^5(c-a)+c^5(a-b)

Asked by  | 14th Dec, 2008, 11:08: AM

Expert Answer:

Problems such as above are solved with the help of cyclic factorisation.

The above is is a cyclic expression. A sample problem has been done below in which you are required to factorise (a-b)^5+(b-c)^5+(c-a)^5.
Let f(a, b, c)=(a-b)^5+(b-c)^5+(c-a)^5. Note that f(a, b, c) = f(a, c, b) = f(b, a, c) = f(b, c, a) = f(c, a, b) = f(c, b, a), i.e. any permutation of a, b and c gives the same expression. Therefore, it is cyclic.

In order to factorize cyclic expressions, we can try to divide the expression by cyclic expressions with lower degrees. For instance, the degree of the expression is 5. If it has non-trivial factors, then some of them must be of degree 1 or 2.

There is only 1 cyclic expression in a, b and c of degree 1, namely \[(a+b+c)\]. By remainder theorem, as f(-b-c, b, c)=(-b-c-b)^5+(b-c)^5+(-b-c-c)^5 neq 0 in general, [(a+b+c) is not a factor of the expression

Next, all cyclic expressions a, b and c of degree 2 are k(a^2+b^2+c^2)+h(ab+bc+ca) for some constants h and k. By some checkings, we know that \[(a^2+b^2+c^2-ab-bc-ca)\] is a factor of the expression. By division, we know that there remains 5(a-b)(b-c)(c-a) when the expression is divided by (a^2+b^2+c^2-ab-bc-ca).

Therefore, (a-b)^5+(b-c)^5+(c-a)^5=5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)

Of course, we may observe that \[f(a,a,c)=0\] to know \[(a-b)\] is a factor of the expression. By its cyclic property, we know that \[(b-c)\] and \[(c-a)\] are factors of the expression. Simple division yields the same result.

 

Above technique is easily applied to given problem to et the required factorisation.


Answered by  | 15th Dec, 2008, 01:26: AM

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