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Let $a,b,c,d$ be real positive numbers. Is there an easy way of seeing that

$$(c-d)(b-d)(b-c)-(c-d)(a-d)(a-c)+(b-d)(a-d)(a-b)-(b-c)(a-c)(a-b)=0 $$

without doing the whole multiplication ?

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  • $\begingroup$ Without expanding the whole expression, you can try extracting one term from each product, check where else it occurs, and determine that it cancels with a negative copy from another product. For example, the first product $(\color{red}c-d)(\color{red}b-d)(\color{red}b-c)$ contributes $c\times b\times b=b^2c$, while the last product $\color{red}{-}(\color{red}b-c)(a\color{red}{-c})(a\color{red}{-b})$ contributes $-b\times(-c)\times(-b)=-b^2c$. $\endgroup$
    – user170231
    Commented Sep 16, 2020 at 21:10
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    $\begingroup$ @user170231: this is the same amount of work as expanding the product. $\endgroup$
    – user65203
    Commented Sep 16, 2020 at 21:20
  • $\begingroup$ This identity is id4_4_1_3a in my list of Special Algebraic Identities. It is a limiting case of a Jacobi elliptic function identity. The identity is valid in any commutative ring. $\endgroup$
    – Somos
    Commented Sep 17, 2020 at 0:06
  • $\begingroup$ @Somos Very interesting, can you expand on the Jacobi comment ? Is there a $n$ variables version that is also connected to Jacobi elliptic function ? $\endgroup$
    – W. Volante
    Commented Sep 17, 2020 at 0:34
  • $\begingroup$ @W.Volante If you read the comments before the id4_4_1_3a identity you will see that is just a special case of a more general identity published by Chamberland and Zeilberger. The next comment is from an article by Glaisher with the Jacobi elliptic sn identity. I don't know if there is a $n$ variable version of that. $\endgroup$
    – Somos
    Commented Sep 17, 2020 at 0:52

3 Answers 3

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Yes! Pick your favorite variable, say $a$, and treat the expression as a polynomial in $a$ (with coefficients that are themselves polynomials in $b,c,d$). It's easy to check that this polynomial has degree at most $2$; but it's also easy to check that it vanishes at $b$, $c$, and $d$ and thus has at least three roots. Therefore it must be the zero polynomial.

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    $\begingroup$ This is an outstanding answer and I wish I could upvote it more than once. $\endgroup$
    – mweiss
    Commented Sep 16, 2020 at 21:26
  • $\begingroup$ So elegant, I love it! $\endgroup$
    – W. Volante
    Commented Sep 17, 2020 at 0:30
  • $\begingroup$ Note that the three roots $b,c,d$ must be distinct if we want to conclude the zero polynomial. $\endgroup$
    – Sil
    Commented Sep 17, 2020 at 1:38
  • $\begingroup$ @Sil: True, but think of $b,c,d$ not as numbers but as actual variables, so that they are automatically distinct. $\endgroup$
    – Greg Martin
    Commented Sep 17, 2020 at 5:27
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    $\begingroup$ The most convenient way to view the polynomial in this context is as an element of $\Bbb R(b,c,d)[a]$, the polynomial ring in one variable over the field $\Bbb R(b,c,d)$. (The fact that $\Bbb R(b,c,d)$ is the field of fractions of another polynomial ring $\Bbb R[b,c,d]$ might appear to muddy the waters, but a field is a field is a field.) $\endgroup$
    – Greg Martin
    Commented Sep 17, 2020 at 7:51
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Write it as$$\begin{align}&\int_a^b(c-d)[2x-c-d]\mathrm{d}x+\int_c^d(a-b)[2y-a-b]\mathrm{d}y\\&=\int_a^b\mathrm{d}x\int_c^d\mathrm{d}y\{2x+2y-a-b-c-d\}.\end{align}$$The symmetry $x\to(a+b) /2-x,\,y\to(c+d)/2-y$ make the rest obvious.

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Making two variables equal makes two products vanish and makes the remaining two cancel each other. E.g. with $a=b$, the equation turns to the identity

$$(c-d)(b-d)(b-c)-(c-d)(b-d)(b-c)\equiv 0.$$ Hence the given expression is of the form

$$P(a,b,c,d)(a-b)(a-c)(a-d)(b-c)(b-d)(c-d).$$

Because of the degrees, $P(a,b,c,d)\equiv 0$.

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