Given:
$$a - \frac1a = b,\qquad b - \frac1b = c, \qquad c - \frac1c = a$$
Find: $\dfrac1{ab} + \dfrac1{bc} + \dfrac1{ac}$
Options: $$a)-3 \qquad b)-1 \qquad c)-6 \qquad d)-9$$
It is based on the image I got from one of my friends. I am getting non-integral value but options are integral.
On
Given $\;a,\;$ set $\;b := a-\frac1a,\; c := b-\frac1b,\; d := c-\frac1c,\; x := a-d.\;$ Factoring expressions gives, $\;x = \frac1a+\frac1b+\frac1c,\quad y := \frac1{ab}+\frac1{bc}+\frac1{ac} = -3 + xa.\;$ Thus, if $\;d=a\;$ then $\;x=0\;$ and $\;y=-3.$
In more detail, $$\;a+b+c = \frac{2-6a^2+3a^4}{a^3-a},\quad ab+ac+bc = abcx = \frac{-1+6a^2-9a^4+3a^6}{a^4-a^2}.$$
On
You can use these equations to get: $$\frac{t^4-3t^2+1}{t^3-t}=a$$ where $t=a-\frac{1}{a}$.
Via a graphing calculator, starting with the principal $a\approx0.5077$, we get $b\approx -1.4619$ and $c\approx -0.7779$. Plugging these into the calculator with that accuracy gave me $-3.000021455$. Not exactly rigorous proof but shows the answer you search for is $-3$
Let $X=\frac1{ab}+\frac 1{ac}+\frac1{bc}$.
By adding the three euations, we find that $$ \tag1\frac1a+\frac 1b+\frac 1c=0$$ hence $$\tag2ab+ac+bc=0.$$ Note that from the firat equation $\frac1{ab}=\frac {a-b}b=\frac ab-1$, and similar for the other equations, so that $$\tag3X=\frac ab+\frac bc+\frac ca-3.$$ We also have from the first equation that $\frac1{ac}=\frac{a-b}c$, and similar for the other equations, so that $$\tag4 X=\frac ac-\frac bc+\frac ba-\frac ca+\frac cb-\frac ab$$ Also, using $(2)$, $$\tag5 0=\left(\frac1{ab}+\frac1{bc}+\frac1{ca}\right)(ab+bc+ac)=3+\frac ac+\frac bc+\frac ab+\frac ba+\frac ca+\frac cb.$$ On the other hand $(4)+2(3)$ says $$3X= \frac ac+\frac bc+\frac ab+\frac ba+\frac ca+\frac cb-6.$$ Together with $(5)$ this gives us $$X=-3.$$