Because Trump Raises Taxes We Will Return to an Old Problem

           I solved a problem on the same topic a while ago. It is an algebraic folklore problem involving Conditional Identities:

          "For non-zero numbers  $x,\;y,\;z$

$$x+y+z=0\;\Leftrightarrow \; \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right )^2 \;"\tag{1}$$

           Indeed,  $x+y+z=0\;\;\underset{xyz \neq 0}{\Leftrightarrow}\;\;\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\;\Leftrightarrow$

$$\Leftrightarrow\;\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\;\Leftrightarrow$$

$$\Leftrightarrow\;\;\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} $$

QED $\square$

          A new problem of this type has appeared.

To better locate it, I only cropped it.

      To solve, we only need to see that the numbers

  $x=a-2b,\;y=2b-3c,\;z=3c-a\;$ check  $x+y+z=0$.

Then we get the answer  $p=\left (\frac{1}{a-2b}+\frac{1}{2b-3c}+\frac{1}{3c-a}\right )^2.$

$\blacksquare$