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We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$

The biggest problem is that the above expression has four variables.I transformed the expression into $$\dfrac{a}{-(a-b)(c-a)(x-a)}+\dfrac{b}{-(b-c)(a-b)(x-b)}+\dfrac{c}{-(c-a)(b-c)(x-c)}$$ and then tried to add them up,but as you can probably guess,it yielded something almost impossible to handle.I tried to factorize it using standard techniques,but that was fruitless too.A hint will be appreciated.

N.B: The chapter deals with Remainder-factor theorem,partial fractions,factoring cyclic homogenous polynomials and manipulating algebraic expressions.

Asaf Karagila
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rah4927
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  • Can you use maple? – mathse May 02 '14 at 12:57
  • @mathse,sure,if you can convince my teacher to allow a laptop during the exam. – rah4927 May 02 '14 at 12:58
  • Why should I want to convice your teacher? – mathse May 02 '14 at 13:00
  • @mathse,why would you you want to use maple to answer a perfectly simple algebra question? – rah4927 May 02 '14 at 13:04
  • Well, apparently it's not simple enough for you to be able to solve it. The reason I would use maple is because the question is mathematically not very challenging. I would also not invert a $4\times 4$ matrix by hand. – mathse May 02 '14 at 13:07
  • @mathse,I used the word "simple" to mean a question which can be understood quite easily.Also,I am sure this has a reasonably elegant solution because it's a textbook question whose neighbouring questions all have reasonably elegant solutions. – rah4927 May 02 '14 at 13:11
  • You have a good start. (I'd move those leading negatives to the numerators.) The next step is to identify the common denominator. Writing each term with that denominator, and then multiplying-out all the numerators will be an ordeal. I cheated and used Mathematica to find the result, so I can tell you: there'll be a lot of cancellation. Considering how very nice the answer is, I'm wondering about the clever way to "see" it in the problem's current form; even so, my initial approach would be brute force: bite the bullet, multiply the numerators out, and watch things cancel. – Blue May 02 '14 at 13:13
  • @rah4927. To be honest, I think someone who wants an answer should be friendlier. So I wouldn't help you even if I could. Good luck with your simplification problem. – mathse May 02 '14 at 13:14
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    @mathse,my first comment(that you apparently seem to have taken as an offense) was intended to be humour.My second comment was induced due to your emphasis on "you". So I am extremely sorry if I seemed to have been unfriendly and rude. – rah4927 May 02 '14 at 13:17
  • The simplification is not difficult. Start with the first two terms. Add them up and simplify. You should get something like $\frac{ab-cx}{(c-a)(c-b)(x-a)(x-b)}$. Then you can use this term with the last one and you get $\frac{x}{(x-a)(x-b)(x-c)}$. – Arash May 02 '14 at 13:18
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    To check, what topic is this homework in? Partial fractions? Remainder factor theorem? Factorization? – Calvin Lin May 02 '14 at 13:35
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    @CalvinLin,all of them.The chapter deals with manipulating and factorizing algebraic expressions.Perhaps I should add that to the body of my question. – rah4927 May 02 '14 at 13:36
  • I believe that the topic you are working on would influence the approach to this problem. I would avoid brute force and think about how to interpret it. I have a solution via partial fractions, think of how to use the rounder factor theorem. – Calvin Lin May 02 '14 at 13:40

3 Answers3

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Observe that the coefficients of 1/(x-a), 1/(x-b), 1/(x-c) seem very cyclic, if not symmetric. This strongly suggests to appeal to the theory of partial fractions. In fact, if in the coefficient of 1/(x-a), we replace the a with x, and do the similar thing for the rest, we will always get

$$ \frac {x} {(x-a)(x-b)(x-c)} $$

Hence, this fraction is equal to your expression through partial fractions. (Yay to avoiding tedious expansions.)

If you are in algebra precalc, it is likely that you have done partial fractions. There is also another solution by the remainder factor theorem.

Calvin Lin
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  • Sorry but I am not entirely sure I understand the first solution.Why does being able to get $\dfrac{x}{(x-a)(x-b)(x-c)}$ by substituting x for $a$ imply that the expression is our answer?I have tried to partially decompose each of the fractions but did not succeed in it. – rah4927 May 02 '14 at 13:47
  • Do you know the quick and dirty method of partial fractions for denominators with linear terms? To find the coefficient of 1/(x-a), You ignore the (x-a), and replace x throughout with a. – Calvin Lin May 02 '14 at 13:50
  • No I indeed did not know that quick and dirty way of PFD.Thanks for letting me know.I will definitely do more research on it apart from just sticking to what school textbooks teach me. – rah4927 May 02 '14 at 14:30
  • When you say that one can use the remainder factor theorem,do you mean the method that Mark Bennet uses in his answer?He uses it to factorize a cyclic expression. – rah4927 May 05 '14 at 05:16
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$$ \frac{a}{(a-b)(a-c)(x-a)}=\frac{a((a-c)-(a-b))}{(b-c)(a-b)(a-c)(x-a)}=\frac{a}{(b-c)(a-b)(x-a)} - \frac{a}{(b-c)(a-c)(x-a)} $$

Thus, $$ \frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-a)(b-c)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}=\frac{1}{(b-c)(b-a)}(\frac{b}{x-b}-\frac{a}{x-a})+ \frac{1}{(c-a)(c-b)}(\frac{c}{x-c}-\frac{a}{x-a}) $$ $$ =\frac{x}{(b-c)(x-a)(x-b)}-\frac{x}{(b-c)(x-a)(x-c)}=\frac{x}{(x-a)(x-b)(x-c)} $$

nature1729
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This answer is to show that this can be done by hand without any particular clever trick, but by being clearheaded and organised. With the original expression being symmetric, you should expect the appearance of symmetric expressions in the expansion, and often a fair amount of cancellation.

If we put everything over a common denominator we get $$\frac {-a(b-c)(x-b)(x-c)-b(c-a)(x-a)(x-c)-c(a-b)(x-a)(x-b)} {(a-b)(b-c)(c-a)(x-a)(x-b)(x-c)}$$

Now we notice that the coefficient of $x^2$ in the numerator is $-a(b-c)-b(c-a)-c(a-b)=0$ and the constant coefficient is $-abc(b-c)-abc(c-a)-abc(a-b)=0$.

The coefficient of $x$ is $$a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)$$$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$

One way to factorise such an expression is to note that it is zero if $a=b$, or $b=c$ or $c=a$ so must have a factor $d(a-b)(b-c)(c-a)$ where d is a constant obtained by equating the coefficient of $ab^2$, for example.

Another way (if you don't have a better idea) is to treat it as a quadratic in $c$ so that $$(b-a)c^2+(a^2-b^2)c+ab(b-a)=(b-a)\left(c^2-(a+b)c+ab\right)=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$$

A further note - the original expression was symmetric. The common denominator had a factor $(a-b)(b-c)(c-a)$ which is skew symmetric, and this can only happen if this same factor, which is the basic skew symmetric expression in three variables, also appears as a factor in the numerator (otherwise the whole expression will fail to be symmetric). The factor has to cancel - and this helps us to know what we are looking for, and to imagine short cuts.

Mark Bennet
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