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Prove that $$\cfrac{1}{a(a-b)(a-c)} +\cfrac{1}{b(b-c)(b-a)} +\cfrac{1}{c(c-a)(c-b)} =\cfrac{1}{abc}$$ for all sets of distinct nonzero numbers $a,b,c $.

Now my question is not about how to solve this but rather why the technique which shows my book works.

Technique:

Rather than showing that the left side equals $\cfrac{1}{abc}$,we show that $$\cfrac{1}{a(a-b)(a-c)} +\cfrac{1}{b(b-c)(b-a)} +\cfrac{1}{c(c-a)(c-b)} -\cfrac{1}{abc}=0 $$

Writing the left side with the common denominator $abc(a-b)(a-c)(b-c)$,we have $$\cfrac{bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)}{abc(a-b)(a-c)(b-c)}=0$$

We can show that this is $0$ by showing that the numerator is $0$.We can do this by looking at the numerator as a polynomial in $c$,meaning let $a$ and $b$ be constants and $c$ be a variable,or $$f(c)=bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)$$

Since $f(c)$ is a quadratic equation ,if we can show that this quadratic has $3$ different roots,then $f(c)=0$ for all $c$.

The proof ends up with showing that $f(a)=0$,$f(b)=0$ and $f(0)=0$.

Now,while I can understand why a quadratic with $3$ roots is the zero polynomial I can't understand why we can treat the numerator as a polynomial and so treat $a,b$ as constants while $c$ as a variable.

Furthermore when we let it be a polynomial we also let $c=a=b$ but the problem in the beginning states that ${a,b,c}$ is all sets of distinct nonzero numbers,so I thought that we can't let $c=a=b$ by definition.

So can someone explain in depth why this is legit to do ?

Mr. Y
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  • This is a good question because it illustrates the difference between different equalities. We have an equality as rational functions and an equality as polynomial functions. The relationship between these equalities is the central part of this question. – Michael Burr Dec 18 '15 at 00:41

2 Answers2

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For a proof that might satisfy a thirst for more symmetry, but which uses a very similar technique, consider the equivalent identity $$\cfrac{bc}{(a-b)(a-c)} +\cfrac{ac}{(b-c)(b-a)} +\cfrac{ab}{(c-a)(c-b)} =1.$$ Let $f(x)$ be the polynomial function defined by $$f(x) = \cfrac{(x-b)(x-c)}{(a-b)(a-c)} +\cfrac{(x-c)(x-a)}{(b-c)(b-a)} +\cfrac{(x-a)(x-b)}{(c-a)(c-b)}.$$ Observe that $f$ is a quadratic polynomial with $f(a)=f(b)=f(c)=1$. It follows that $f(x)-1$ is a quadratic with three roots, so $f(x)-1=0$ identically. Now compare constant terms of the identity $f(x)=1$.

Dustan Levenstein
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  • I don't know if my conclusion from your answer is correct. We have that $g(x)=f(x)-1$ is $0$ for any $x$ ,from that it follows that it applies to any value of $a,b,c$ because we can consider $b=(x-b_1)$,$a=(x-a_1)$, $c=(x-c_1)$ where $a_1,b_1,c_1$ are constants . – Mr. Y Dec 17 '15 at 17:49
  • I'm not really clear on what you're trying to say? In my proof, $a, b, c$ are any fixed choice of three distinct numbers. The same analysis applies equally well to any other choice, though. – Dustan Levenstein Dec 17 '15 at 18:44
  • I was trying to say that since we have proved it for a fixed choice of three distint numbers $a,b,c$ it can be applied then to any other number given that we can get any other $a$ or $b$ or $c$ from the fixed value we have given them in the beginning,for example any $a$ can be expressed as $x-a_1$ where $a_1$ is some fixed value we have given in the beginning to transform our expression in a polynomial.I was asking if this interpretation is correct. – Mr. Y Dec 17 '15 at 19:00
  • Oh I see. Well this will allow you to replace $(a, b, c)$ with $(a+t, b+t, c+t)$, but that's not general enough. To get the general result, just observe that the analysis does not depend on the particular choice of $a, b, c$, as long as they are distinct. – Dustan Levenstein Dec 17 '15 at 19:02
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First, I'll mention an elementary way to finish the proof. Observe that this could be done by expanding the following product: $$ f(c)=bc(b−c)−ac(a−c)+ab(a−b)−(a−b)(a−c)(b−c). $$ At the end, you should get everything to cancel.

Instead, the proof uses a tricky method. By inspection, we can see that $$ bc(b−c)−ac(a−c)+ab(a−b)−(a−b)(a−c)(b−c) $$ can be written as $$ \text{Something}\cdot c^2+\text{Something}\cdot c+\text{Something} $$ where these "Somethings" are written in terms of $a$ and $b$. The goal of the proof is to show that these "Somethings" are always zero (independent of the choice of $a$ or $b$).

Now, all of $a$, $b$, and $c$ are variables, but we can choose to pick (arbitrary) values for $a$ and $b$ and not change them. Then, the quadratic above becomes an actual quadratic (all the coefficients are numbers after substituting in for $a$ and $b$). At that point, we are free to choose $c$ as we wish. By finding three distinct values where $c$ vanishes, the quadratic then has zero coefficients, and this holds for any choice of $a$ and $b$.

It is somewhat more tricky to explain why we can drop the distinct assumption without some experience with polynomials. When a polynomial is equal to $0$ at all but a small number of points (I can make this more precise, but it seems beyond the tags for the OP). Then, when a polynomial (the numerator) is equal to $0$ at all but the points where $a=b$, $a=c$, or $b=c$. We can extend it to all possible values of $a$, $b$, and $c$. Basically, a polynomial in one variable either has finitely many roots or it's zero (there are no other options). So, if a polynomial is $0$ at all but the points where $a=b$, $a=c$, or $b=c$, it already has too many roots to be nonzero.

The only reason that the original problem required $a$, $b$, and $c$ to be distinct was so that the fractions didn't have $0$ in the denominator. If you use limits from calculus, you may be able to get the original equality to "hold" even when $a$, $b$, or $c$ are equal.

Michael Burr
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    It is given that $a,b,c$ are distinct so how can you conclude they have them as zeroes? – user19405892 Dec 17 '15 at 16:40
  • So when we treat the numerator as a polynomial the set constraint for $a,b,c$ isn't important anymore for the reason you've indicated? – Mr. Y Dec 17 '15 at 16:52
  • @Mr.Y That pretty much sums it up. To answer the question fully requires classical algebraic geometry and that's not appropriate for this type of question. – Michael Burr Dec 17 '15 at 16:57