# CTK Insights

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02 Jul

### An Engaging Algebraic Identity

A question has been asked on a linkedin group to prove the following engaging identity

Prove

$(\frac{b-c}{a} + \frac{c-a}{b}+ \frac{a-b}{c})(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})=9,$

provided $a+b+c=0$.

One of the posts pointed to a solution at Stevens Society of Mathematicians. What follows is a slight simplification of that proof.

Denote the left factor $L(a,b,c)$ and the right factor $R(a,b,c)$. Observe that whenever two of the arguments in $R$ are equal, the whole expression vanishes. For example,

\begin{align}L(a,a,c)&=\frac{a-c}{a}+\frac{c-a}{a}+\frac{a-a}{c} \\ &=\frac{a-c}{a}-\frac{a-c}{a}=0.\end{align}

Adding the fractions in $L(a,b,c)$, $L(a,b,c)=\frac{L'(a,b,c)}{abc}$. What we just showed implies that the numerator $L'$ is divisible by $(a-b)(b-c)(c-a)$. Multiplying through confirms that $L'(a,b,c)=-(a-b)(b-c)(c-a)$.

Now, let's turn to the right factor. Up to now we have not used the condition $a+b+c=0$. It's time we do. Introduce

$\begin{cases}x = b - c \\ y = c - a \\ z = a - b.\end{cases}$

Seen as a system of linear equations with $a,b,c$ as unknown, it's degenerate because $x+y+z=0$. The situation improves if we replace any of the equations with $a+b+c=0$. Then, for example,

\begin{align}y-z &= (c-a)-(a-b) \\ &= (b+c)-2a \\ &=-3a. \end{align}

Similarly, $-3b=z-x$ and $-3c=x-y$. This allows us to express the right factor $R(a,b,c)$ in terms of $x,y,z$:

$R(a,b,c)=-\frac{1}{3} (\frac {y-z} {x} + \frac {z-x} {y} + \frac {x-y} {z} ).$

This is exactly the same form as $L(a,b,c)$, implying that

$R(a,b,c)=\frac{1}{3}\frac{(y-z)(z-x)(x-y)}{xyz}=-9\frac{abc}{(b-c)(c-a)(a-b)}$.

Finally, $L(a,b,c) \cdot R(a,b,c) = 9.$