JBMO 2020 problem 1.

@above: Note that $a=1, b=42, c=\frac{1}{42}$ clearly also is a solution.

Extension

Squaring the first equation we get \[ a^2+b^2+c^2+2(ab+bc+ca)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}. \]This, together with $ab+bc+ca=abc(a+b+c)$, and the second equation, then yields \[ 2abc^2(a+b+c)=2(a+b+c)\Rightarrow ((abc)^2-1)(a+b+c)=0. \]Now if $a+b+c=0$, we get $ab+bc+ca=0$ as well, which in turn yields $a=b=c=0$, which is clearly not possible. Hence, $abc=\pm 1$. Notice, furthermore, that if $(a,b,c)$ is a solution so is $(-a,-b,-c)$. For this reason, we will assume wlog $abc=1$ in the remainder. Now, we have $a+b+c=ab+bc+ca$ and $abc=1$ (in particular, it is easily seen these two conditions yield automatically the second equation). Setting $c=\frac{1}{ab}$, we have \[ ab(a+b)+1=(ab)^2+(a+b). \]Let $m=ab$ and $n=a+b$. In terms of this, we have the quadratic (in $m$): $m^2-mn+n-1=0$, which factors as $(m-n+1)(m-1)=0$, which translates \[ (ab-a-b+1)(ab-1)=0. \]If $ab=1$, we get the family $(x,\frac1x,1)$ (and its permutations, with sign). If $ab-a-b+1=0$ then either $a=1$ or $b=1$. If $a=1$, then any $b$ works, which gives $c=\frac1b$, which is again included in the aforementioned family. This concludes the solution.

dangerousliri wrote: Find all triples (a,b,c) of real numbers such that the following system holds: $$\begin{cases} a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{cases}$$ Proposed by Dorlir Ahmeti, Albania Nice problem! Kudos to dangerousliri for proposing such fun-to-solve problems in olympiads. Posted a solution with a bit of motivation and thought process on my blog here

For this part a+b+c=ab+ac+bc abc-ab-ac-bc+a+b+c-1=0 (a-1)(b-1)(c-1)=0 One of a,b or 3 must be equal to 1.Let suppose c=1 then ab=1 b=1/a So (1,X,1/X) and all permutations are solution Now for abc=-1 then a+b+c=-(ab+ac+bc) abc+ab+ac+bc+a+b+c+1=0 (a+1)(b+1)(c+1)=0 So one of a,b or c must be equal to -1 Let suppose c=-1 Then ab=1 b=1/a So (-1,X,1/X) and all permutations are solution

completing the square in equation (2) and using equation (1) it follows that $\sum ab=\sum \frac{1}{ab} \implies (abc)^2\sum \frac{ab}{abc}=\sum \frac{abc}{ab} \implies (abc)^2=\pm 1$ Now,considering two polynomials with roots respectively ${a,b,c}$ and ${1/a,1/b,1/c}$ it follows that $\{a,b,c\}=\{1/a,1/b,1/c\}$ This is also suffiicient.

Let $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=k,$ where $k\in\mathbb{R},$ and $abc\neq 0,$ we have $a+b+c=k$ and $ab+bc+ca=kabc.$ Then, under the condition $abc\neq 0,$ $a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ $\Longleftrightarrow (a+b+c)^2-2(ab+bc+ca)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$ $\Longleftrightarrow k^2-2(ab+bc+ca)=k^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$ $\Longleftrightarrow ab+bc+ca=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$ $\Longleftrightarrow kabc=\frac{k}{abc}$ $\Longleftrightarrow k(abc+1)(abc-1)=0$ and $abc\neq 0.$ Case 1 : $k=0$ From $a+b+c=ab+bc+ca=0,$ we have $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0$, since $a,\ b,\ c$ are real numbers, we obtain $a=b=c=0,$ which contradicts $abc\neq0.$ Case 2 : $abc=-1$ From $a+b+c=k$ and $ab+bc+ca=kabc$, by Vieta, $a,\ b,\ c$ are the real solutions to the cubic equation $x^3-kx^2-kx+1=0$ $\Longleftrightarrow (x+1)\{x^2-(k+1)x+1\}=0.$ $\Longleftrightarrow x = -1,\ x^2-(k+1)x+1=0.$ Remark that by Vieta , the product of two solutions to the quadratic equation $x^2-(k+1)x+1=0$ is one, hereby we can set the solutions as $t, \frac{1}{t}.$ Then, by Vieta again, the sum of these solutions $t+\frac{1}{t} = k+1.$ Now, the necessary and sufficient condition for which the quadratic equation $x^2-(k+1)x+1=0$ has real solution(s) is given by $D=(k+1)^2 - 4 \geq 0\Longleftrightarrow |k+1| \geq 2$, which is guaranteed by the following reason : Since the product of the solutions $t$ and $\frac{1}{t}$ is one, or a positive real number, thus both of the sign of $t$ and $\frac{1}{t}$ would coincide, hereby, $\left|t+\frac{1}{t}\right|= |t| + \left|\frac{1}{t}\right|\geq 2$ by the AM-GM inequality, yielding $|k+1|\geq 2.$ Therefore, for nonzero real number $t$, we obtain $\{a, b, c\} = \{-1,\ t,\ \frac{1}{t}\}$ with some real number $t\neq 0.$ Case 3 : $abc=1$ From $a+b+c=k$ and $ab+bc+ca=kabc$, by Vieta, $a,\ b,\ c$ are the real solutions to the cubic equation $x^3-kx^2+kx-1=0$ $\Longleftrightarrow (x-1)\{x^2-(k-1)x+1\}=0.$ $\Longleftrightarrow x = 1,\ x^2-(k-1)x+1=0.$ Remark that by Vieta , the product of two solutions to the quadratic equation $x^2-(k-1)x+1=0$ is one, hereby we can set the solutions as $u, \frac{1}{u}.$ Then, by Vieta again, the sum of these solutions $t+\frac{1}{t} = k-1.$ Now, the necessary and sufficient condition for which the quadratic equation $x^2-(k-1)x+1=0$ has real solution(s) is given by $D=(k-1)^2 - 4 \geq 0\Longleftrightarrow |k-1| \geq 2$, which is guaranteed by the following reason : Since the product of the solutions $u$ and $\frac{1}{u}$ is one, or a positive real number, thus both of the sign of $u$ and $\frac{1}{u}$ would coincide, hereby, $\left|u+\frac{1}{u}\right|= |u| + \left|\frac{1}{u}\right|\geq 2$ by the AM-GM inequality, yielding $|k-1|\geq 2.$ Therefore, for nonzero real number $t$, we obtain $\{a, b, c\} = \{1,\ u,\ \frac{1}{u}\}$ with some real number $u\neq 0.$ Consequently, the desired triplets $(a,\ b,\ c)$ of nonzero real numbers are $\{a, b, c\} = \{\pm 1,\ ,\lambda,\ \frac{1}{\lambda}\}$ with some real number $\lambda\neq 0.$

$\definecolor{A}{RGB}{0,148,79}\color{A}\fbox{Identity 1.}$ $$\definecolor{A}{RGB}{80,0,200}\color{A}(a^2+b^2+c^2)(ab+bc+ca)^2-(a^2b^2+b^2c^2+c^2a^2)(a+b+c)^2=-2(ab-c^2)(bc-a^2)(ca-b^2)$$ $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ Because $a,b,c\neq 0$ we get from $\definecolor{A}{RGB}{0,148,79}\color{A}\text{identity 1.}$ the right side, while left side is from the system of equations: $$0=(a^2+b^2+c^2)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^2-\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)(a+b+c)^2=-\frac{2}{a^2b^2c^2}\cdot(ab-c^2)(bc-a^2)(ca-b^2)$$Equations are symmetric, so assume WLOG $c^2=ab$ (hence immediately $ab>0$). The second equation in #1 becomes $$a^2+b^2+ab=\frac{a^2+ab+b^2}{a^2b^2}.$$Observe $a^2+ab+b^2=\frac12(a^2+b^2+(a+b)^2)>0$ hence $$a^2b^2=1.$$$$ab=1\wedge c^2=1.$$ $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Answer}$ Direct substitution proves that triple $$\definecolor{A}{RGB}{235,79,150}\color{A}(a,b,c)=\left(x,\frac1{x}, \pm 1\right)$$and all the permutations work for all $\definecolor{A}{RGB}{235,79,150}\color{A}x\in\mathbb{R}\setminus\lbrace0\rbrace.\blacksquare$ #1758

Claim. The answer is $\boxed{(\pm 1,x,\frac{1}{x})}$ and permutations. Proof. Squaring the first equation, substracting the second and dividing by two we get $$\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=bc+ca+ab$$Taking into the account this equality and the first given we get that $$\frac{a+b+c}{bc+ca+ab}=\frac{1}{abc}=abc$$So $abc=\pm 1$. Case 1. $abc=1$ Then $c=\frac{1}{ab}$ and substituting into the first equation we get $$(ab-1)(a-1)(b-1)=0$$$ab=1 \implies c=1$, thus WLOG $a=1$. Thus $c=\frac{1}{b}$. This triple and its permutations certainly work. Case 2. $abc=-1$ Then $c=-\frac{1}{ab}$ and substituting into the first equation we get $$(ab-1)(a+1)(b+1)=0$$$ab=1 \implies c=-1$, thus WLOG $a=-1$. Thus $c=\frac{1}{b}$. This triple and its permutations certainly work. Thus, the answer is $(\pm 1,x,\frac{1}{x})$ and permutations.

For storage.

RMOAspirantFaraz wrote: $abc=\pm1$ Hence the ordered pair of solution is of the form $(a,b,c)=\left(\pm 1,k,\frac{1}{k}\right)$ You've left out at least half of the work.

Squaring the first equation and subtracting the second gives: $$ab+bc+ca=\frac1{ab}+\frac1{bc}+\frac1{ca}$$Then $abc(ab+bc+ca)=a+b+c=\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}$, so either $ab+bc+ca=0$, $abc=-1$, or $abc=1$. Since $(a,b,c)$ is a solution iff $(-a,-b,-c)$ is, we can ignore the case $abc=-1$. Case 1: $ab+bc+ca=0$ Then: $$\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}=\frac1{a^2}+\frac1{b^2}+\frac1{c^2}=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2=a^2b^2c^2(ab+bc+ca)^2=0$$so $a^2b^2+b^2c^2+c^2a^2=0$. But by the trivial inequality $ab=bc=ca=0$, so some of $a,b,c$ are zero which is impossible. Case 2: $abc=1$ Substituting $c=\frac1{ab}$ into the first equation, we have: $$a+b+\frac1{ab}=\frac1a+\frac1b+ab\Leftrightarrow a^2b+ab^2+1=a+b+a^2b^2\Leftrightarrow(ab-1)(a-1)(b-1)=0\Leftrightarrow(a-1)(b-1)(c-1)=0.$$WLOG $c=1$, then $ab=1$ too. We can check that any permutation of $\boxed{\left(x,\frac1x,1\right)}$ works for $x\in\mathbb R\setminus\{0\}$. The solutions for $(-a,-b,-c)$ are permutations of $\boxed{\left(-x,-\frac1x,-1\right)}$.

Squaring the first equation yields $$a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{2}{ab} + \frac{2}{bc} + \frac{2}{ac}.$$which implies $$ab + bc + ac = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}.$$Now multiply by $abc$ on both sides to get $$abc(ab + bc + ac) = a + b + c. \ \ \ \ (*)$$Multiplying the first equation by $abc$ gives $$abc(a+b+c) = ab + bc + ab.$$Combining with the previous equation, $$abc(a+b+c) = abcabc(ab + bc + ac) = ab + bc + ac \implies abc = \pm 1.$$Substituting this to $(*)$ gives $\pm (a+b+c) = ab + bc + ac = u$. By Vieta's, if $abc = 1$, $a,b,c$ are roots to the equation $x^3 - ux^2 + ux - 1 =0$. We easily see that $x=1$ is a solution. WLOG let $a=1$, then $bc = 1 \Longleftrightarrow b = \frac{1}{c}$. We see that this satisfies the equations. Hence $(1,\frac{1}{n},n)$ in some permutation is a solution. Similarly, if $abc = -1$, $a,b,c$ are roots to the equation $x^3 + u x^2 + ux + 1 = 0$ in which $-1$ is a solution. Hence WLOG let $a = -1$, then $bc = 1$. We see that this also satisfies to given equations. Hence the solutions are $(a,b,c) = (\pm 1 , \frac{1}{n},n)$ in some permutation and real number $n \not = 0$.

$$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =\frac{ab+bc+ca}{abc}$$$$a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \iff ab+bc+ca=\frac{a+b+c}{abc}$$Thus either $abc\in \{1,-1\}$ or $$a+b+c=ab+bc+ca=0\iff a^2+b^2+c^2=0$$absurd. Hence if $abc=1$ then $$a+b+c=ab+bc+ca\iff (a-1)(b-1)(c-1)=0$$thus $$\{a,b,c\}=\left \{1,n,\frac{1}{n}\right \}$$for any $n$ and this works. The case $abc=-1$ is analogous.

my solution (which is also mathmax12's) solution: Squaring this we have $a^2+b^2+c^2+2(ab+ac+bc)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc},$ now we also have $ab+ac+bc=abc(a+b+c)$, so $((abc)^2-1)(a+b+c)=0$, so $abc=1$ or $abc=-1$, now we have that $(ab-a-b+1)(ab-1)=0.$ When $ab=1$. We have $\boxed{(x, \frac{1}{x}, 1)}$, and its permutations with sign.

we have $a+b+c=\frac{ab+bc+ac}{abc}$ and $a^2+b^2+c^2 =\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}$ Also $a^2+b^2+c^2+2(ab+bc+ca)=\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}+\frac{2(a+b+c)}{abc}$ so we have $\frac{(a+b+c)}{abc}=ab+bc+ca \implies \frac{ab+bc+ca}{a^2b^2c^2}=ab+bc+ca \implies$ either $ab+bc+ca=0$ or $abc=\pm 1$ Claim:- $ab+bc+ca \neq 0$ Pf:- FTSOC assume the contrary then we have $ab+bc+ca =0 \implies a^2+b^2+c^2=0$ which implies all of $a,b,c$ is $0$ which is absurd. Hence this is a contradiction and claim follows. $\square$ Now we have $abc= \pm 1$ which gives $\left(\beta , \frac{1}{\beta}, 1\right) , \left(-\beta, \frac{-1}{\beta},1 \right)$ and their permutations to be working triples for every $\beta \in \mathbb{R} \setminus \{0\}$

a^2+b^2+c^2=(a+b+c)^2 - 2(ab+bc+ac)=(1/a+1/b+1/c)^2 - 2(ab+bc+ac)= (1/a^2+1/b^2+1/c^2) + 2(1/ab+1/bc+1/ac - ab - bc - ac) => 1/ab+1/bc+1/ac - ab - bc - ac= 0 => a+b+c = (ab+bc+ac)abc => 1/a+1/b+1/c =(ab+bc+ac)abc => ab+bc+ac=(ab+bc+ac)(abc)^2 => abc = 1 We go back to the equation: 1/ab+1/bc+1/ac - ab - bc - ac= 0 1/ab=1/abc/c=1/1/c=c => a+b+c-ab-bc-ac=0 c=abc/ab = 1/ab => a+b+c=ab+bc+ac => a+b+1/ab=ab+ 1/a+1/b => (a+b)ab+1=ab^2+(a+b) (a+b)(ab-1)=ab^2-1 (a+b)(ab-1)=(ab-1)(ab+1) (ab-a-b+1)(ab-1)=0 Case1 ab=1 =>c=1 (a,b,c)=(a,1/a,1) (not necessarily in this order) and by trial and error we see that it works Case 2 ab-a-b=-1 => (a-1)(b-1)=0 and we get the same solution Tell me if I have mistakes

If we find square of first equation; a²+b²+c²+2ab+2ac+2bc=1/a²+1/b²+1/c²+2/ab+2/ac+2/bc Instead of a²+b²+c², we can write 1/a²+1/b²+1/c². Let's do: 2ab+2ac+2bc=2/ab+2/ac+2/bc Dividing the sides into 2; ab+ac+bc=1/ab+1/ac+1/bc This have a solution (1,1,1) And also (x; 1/x; 1)

I have very clown writeup but will post anyways We claim the answer is $ \boxed{(a,b,c)=\left(-1, b ,\frac{1}{b}\right), \text{ }\left(a, -1, \frac{1}{a}\right), \left(a, \frac{1}{a}, -1\right), \text{ } \left(1,b,\frac{1}{b}\right), \text{ }\left(a,\frac{1}{a},1\right),\text{ and }\left(a,1,\frac{1}{a}\right)}$ for all real numbers $a,b\neq 0.$ These clearly work. We proceed as follows. Let $a+b+c=x,$ $ab+bc+ac=y,$ and $abc=z.$ We have $a+b+c=x=\frac{ab+bc+ac}{abc}=\frac{y}{z},$ so $xz=y.$ Also, $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=x^2-2y =\frac{a^2b^2+b^2c^2+a^2c^2}{a^2b^2c^2}=\frac{(ab+bc+ac)^2-2(abc)(a+b+c)}{z^2}=\frac{y^2-2xz}{z^2}=\frac{y^2-2y}{z^2}.$ As such, $y^2-2y=x^2z^2-2yz^2 =y^2-2yz^2,$ so $yz^2=y.$ As such, $y=0,$ or $z=\pm 1.$ Note that if $y=0,$ then $x=\frac{y}{z}=0,$ so by vieta's, $a,b,$ and $c$ are roots of the cubic $d^3-xd^2+yd-z=d^3-z=0.$ However, note that this cubic only has one real root for $z\neq 0,$ at $\sqrt[3]{z},$ and if $z=0,$ then $(a,b,c)=(0,0,0)$ which is clearly not a solution, so there exists no real solutions when $y=0.$ As such, $z=\pm1.$ If $z=1,$ $y=z,$ so $a,$ $b,$ and $c$ are roots of the cubic $d^3-xd^2+yd-z=d^3-xd^2+xd-1=0.$ Note that $d=1$ is a root of this cubic, so at least one of $a,$ $b,$ and $c$ is $1.$ WLOG, $a=1.$ We have $1+b+c=1+\frac{1}{b}+\frac{1}{c},$ so $b+c=\frac{b+c}{bc},$ so $b+c=0,$ or $bc=1.$ Also, $1+b^2+c^2=1+\frac{1}{b^2}+\frac{1}{c^2},$ so $b^2+c^2=\frac{b^2+c^2}{b^2c^2},$ so $b^2+c^2=0$ or $b^2c^2=1.$ Clearly, $b^2+c^2\neq 0,$ so $bc=\pm 1.$ If $bc=1,$ we have $(a,b,c)=\left(1,b,\frac{1}{b}\right),$ which works for all $b\neq 0,$ so $(a,b,c)=\left(1,b,\frac{1}{b}\right), \text{ }\left(a,\frac{1}{a},1\right),\text{ and }\left(a,1,\frac{1}{a}\right)$ work for any real numbers $a,b\neq 0.$ If $bc=-1,$ then $bc\neq 1,$ so $b+c=0,$ which yields $(b,c)=(1,-1),$ or $(b,c)=(-1,1),$ which yields the solutions $(a,b,c)=(1,1,-1), \text{ } (1,-1,1), \text{ and }(-1,1,1),$ which all work. If $yz=-1,$ then $x=-y.$ Note that $a,$ $b,$ and $c$ are roots of the cubic $d^3-xd^2+yd-z=d^3-xd^2-xd+1=0.$ Note that $-1$ is a root of the cubic, so at least $1$ of $a,b,$ and $c$ is $-1.$ WLOG, let $a=-1.$ We have $b+c=\frac{b+c}{bc},$ and $b^2+c^2=\frac{b^2+c^2}{b^2c^2},$ so $b+c=0$ or $bc=1,$ and $b^2+c^2=0$ or $b^2c^2=1.$ Note that $b^2+c^2\neq 0,$ so $b^2c^2=1,$ so $bc=\pm1.$ If $bc=1,$ we have the solutions $(a,b,c)=\left(-1, b, \frac{1}{b}\right), \text{ }\left(a, -1, \frac{1}{a}\right), \text{ and }\left(a, \frac{1}{a}, -1\right)$ for real numbers $a,b\neq 0,$ which all work. If $bc=-1,$ then $b+c=0,$ yielding the solutions $(a,b,c)=(-1,-1,1),\text{ }(-1,1,-1), \text{ and }(1,-1,-1).$ Note that the solutions $(a,b,c)=(-1,-1,1),\text{ }(-1,1,-1), \text{ and }(1,-1,-1)$ are a special case of the solutions $(a,b,c)=\left(1,b,\frac{1}{b}\right), \text{ }\left(a,\frac{1}{a},1\right),\text{ and }\left(a,1,\frac{1}{a}\right)$ for all reals $a,b\neq 0,$ and the solutions $(a,b,c)=(1,1,-1), \text{ } (1,-1,1), \text{ and }(-1,1,1),$ are a special case of the solutions $(a,b,c)=\left(-1, b, \frac{1}{b}\right), \text{ }\left(a, -1, \frac{1}{a}\right), \text{ and }\left(a, \frac{1}{a}, -1\right)$ for all real numbers $a,b\neq 0.$ As such, the only solutions are $ (a,b,c)=\left(-1, b ,\frac{1}{b}\right), \text{ }\left(a, -1, \frac{1}{a}\right), \left(a, \frac{1}{a}, -1\right), \text{ } \left(1,b,\frac{1}{b}\right), \text{ }\left(a,\frac{1}{a},1\right),\text{ and }\left(a,1,\frac{1}{a}\right)$ for all real numbers $a,b\neq 0,$ as desired. $\Box$

Consider the polynomial with roots $a, b, c$. Sorry for the abuse of notation but let the coefficients be $x^3+ax^2+bx+c$ where the $a, b, c$ here have nothing to do with the problem im just too lazy to comphrehend my scratch work and change it. Change the $a, b, c$ in the problem to $x, y, z$. Then the first condition translates to $ac=b$ and the second into $a^2-2b=\frac{b^2-2ac}{c^2} = \frac{b^2-2b}{c^2} \implies b=bc^2$ after substituiting for $a$. If $b=0$, then we quickly find that $a=0$ which translates the cubic into one with roots of unity, impossible. Then $c=\pm 1$. If $c=1$, then $a=b$. The cubic becomes $x^3+ax^2+ax+1$ which has a root of $-1$ and the other two roots must be reciprocal. If $c=-1$, then $a=-b$ and the cubic is $x^3+ax^2-ax-1$. A root is $-1$ and the other two are also reciprocal, so the solutions are of the form $\boxed{\left( 1, x, \frac{1}{x} \right)}$ up obviously to permutations and signs.