Express $(a+b+c)abc$ purely in terms of $d$, given $a^2-\frac{1}{b}=b^2-\frac{1}{c}=c^2-\frac{1}{a}=d$

Question

Given that $a \ne b \ne c $ are non zero reals satisfying $$a^2-\frac{1}{b}=b^2-\frac{1}{c}=c^2-\frac{1}{a}=d \tag {*}$$ Then express $(a+b+c)abc$ purely as a function of $d$.

My hard effort:

Taking pair wise subtractions $$ a^{2}-\frac{1} {b}=b^{2}-\frac{1} {c}=c^{2}-\frac{1} {a} $$ $$ \Rightarrow( a-b ) ( a+b )=\frac{c-b} {b c} $$ $$ ( b-c ) ( b+c )=\frac{a-c} {a c} $$ $$ ( a-c ) ( a+c )=\frac{a-b} {a b} $$ $$ \Rightarrow( a+b ) ( b+c ) ( c+a )=\frac{-1} {a^{2} b^{2} c^{2}} \tag {1} $$ Now I assumed $a+b+c=p, ab+bc+ca=q, abc=r$. So $(1)$ becomes $$pq-r=\frac{-1}{r^2} \tag {2}$$ Also from $(*)$ adding all equations with $d$, we get $$ a^{2}+b^{2}+c^{2}-\frac{1} {a}-\frac{1} {b}-\frac{1} {c}=3 d $$ $$ \Rightarrow( a+b+c )^{2}-2 ( a b+b c+c a )-\frac{( a b+b c+c a )}{abc}=3 d $$ $$ \Rightarrow p^{2}-2 q-\frac{q} {r}=3 d \tag {3} $$ Now I eliminated $q$ from $(2),(3)$, so we get $$2r+1-\frac{2r+1}{r^3}=p^3-3pd \tag {4}$$ Any help from here?

Answer 1

Here is a by-hand solution. As in the question body, let $p=a+b+c$, $q=ab+bc+ca$, and $r=abc$. Our aim is to find $pr$.

We can compute $$qd=\sum_{\mathrm{cyc}}bcd=\sum_{\mathrm{cyc}}bc\left(a^2-\frac1b\right)=\sum_{\mathrm{cyc}}(a^2bc-c)=p(r-1).\tag{$5$}$$ We first assume $r\not\in\{1,-1/2\}$; we will fix this up later. Combining this with identity (2) in the question body, i.e. $pq=r-1/r^2$, gives $$d(r^3-1)=pqr^2d=pr^2p(r-1),$$ so $(pr)^2=d(r^2+r+1)$. We now use the other identity (3) discovered in the question body: $p^2r-q(2r+1)=3dr$. Multiplying by $r$ gives $$qr(2r+1) = p^2r^2 - 3dr^2 = d(-2r^2+r+1)=-d(2r+1)(r-1).$$ Since $r\neq-1/2$, we conclude $$-d^2(r-1) = dqr(2r+1)=p(r-1)r;$$ since $r\neq 1$, we get $pr=-d^2$, as desired.

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It remains to treat the degenerate cases $r=1$ and $r=-1/2$. For this we will use an additional sum: $$(p^2-2q)d=\sum_{\mathrm{cyc}}b^2d=\sum_{\mathrm{cyc}}b^2\left(a^2-\frac1b\right)=q^2-2pr-p.$$ (The last equality is by expanding out $q^2$.) Adding twice (5) gives $$p^2d=q^2-3p.\tag{$6$}$$

• If $r=1$, then we have by (5) that $qd=0$. So either $q=0$ or $p=d=0$. If $d=0$ then $a^8=b^4=c^2=a$, and so $a=1$, meaning $b=c=1$, contradicting distinctness. So $q=0$. Then (3) gives $p^2=3d$ and (6) gives $p^2d=-3p$. These are enough to give $pr=p=-d^2$.

• If $r=-1/2$, then (2) gives $pq=-9/2$, (3) gives $p^2=3d$, and (6) gives $p^2d+3p=q^2$. Using (2) to solve for $q$, (3) to solve for $d$, and plugging them into (6) gives $$\frac{p^4}3+3p=p^2d+3p=q^2=\frac{81}{4p^2}.$$ Simplifying gives $p^6+9p^3=243/4$, which factors as $(p^3+27/2)(p^3-9/2)=0$. If $p^3=9/2$, then $$-18pr=9p=2p^4=18d^2,$$ and so $pr=-d^2$ holds. Otherwise, $p=-3/\sqrt[3]{2}$. Then $q=3/\sqrt[3]{4}$, and so the cubic with roots $a$, $b$, and $c$ is $$x^3-\frac3{\sqrt[3]2}x^2+\frac3{\sqrt[3]4}x-\frac12.$$ This factors as $(x-\sqrt[3]{1/2})^3$, and so $a=b=c$ in this case. This finishes the proof.

Answer 2

You have three equations in three unknowns $(a,b,c)$, call this system $S$,

$$a^2-\frac{1}{b}=d\\ b^2-\frac{1}{c}=d\\ c^2-\frac{1}{a}=d$$

so can be solved where $a\neq b\neq c$. However, since this is a non-linear system, one can expect that $(a,b,c)$ are generally roots of non-linear equations as well.

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I. Sextic solution

Using resultants, your $(a,b,c)=(x_1,\,x_2,\,x_3)$ are three roots (correctly ordered) of the sextic with a solvable Galois group,

$$(1 - d^3) x^6 - d^2 x^5 - 3d(1- d^3) x^4 + (1 + 2 d^3) x^3 + d^2(4 - 3 d^3) x^2 - d^4 x + (1- d^3)^2 = 0$$

with discriminant $D = ( \color{blue}{4 d^3 - 3})^3 (16 d^6 - 36 d^3 + 27)^2$. However, the sextic factors over a square root extension and we get two cubics,

$$x^3+\frac{d^2\big(1-\sqrt\beta\big)}{2(d^3-1)}x^2+\frac{d\big(3-2d^3-\sqrt\beta\big)}{2(d^3-1)}x+\frac{\big(1+\sqrt\beta\big)}{2}=0$$

for $\pm\sqrt{\beta}$ and where $\beta = \color{blue}{4 d^3 - 3}$. By the elementary symmetric polynomials, we have,

\begin{align} x_1+x_2+x_3 &= -\frac{d^2\big(1-\sqrt\beta\big)}{2(d^3-1)}\\[5pt] x_1x_2x_3 &=-\frac{\big(1+\sqrt\beta\big)}{2}\end{align}

where the product of both removes the square root and simplifies to,

$$abc(a +b + c) = x_1x_2x_3(x_1+x_2+x_3) = x_4x_5x_6(x_4+x_5+x_6) =\color{red}{-d^2}$$

which is essentially the same relation found by cybcat, Calvin Lin, and Dietrich Burde in a deleted answer. However, since the sextic has six roots, then for a given $d$, there are generally two triplets $(a,b,c)$.

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II. Example

Let $\color{red}{d=2}$ and we get the sextic,

$$-7 x^6 - 4 x^5 + 42 x^4 + 17 x^3 - 80 x^2 - 16 x + 49=0$$

with one factor as,

$$14x^3+4(1+\sqrt{29})x^2-2(13-\sqrt{29})+7(1-\sqrt{29})=0$$

and the other the negative case of $\pm\sqrt{29}$. We arrange the roots of the first cubic as,

\begin{align} x_1 &\approx -1.711333828766285769348878\\ x_2 &\approx 1.0768163371956621036573432\\ x_3 &\approx -1.189815310467806057522953\end{align}

and the second cubic as,

\begin{align} x_4 &\approx 1.1742622838169903974214464\\ x_5 &\approx -1.610025723419787135772065\\ x_6 &\approx 1.6886676702126550329936787 \end{align}

Then,

$$abc(a +b + c) = x_1x_2x_3(x_1+x_2+x_3) = x_4x_5x_6(x_4+x_5+x_6) =\color{red}{-2^2}$$

Of course, these two triplets $x_k$ also satisfy the system $S$,

$$a^2-\frac{1}{b}=b^2-\frac{1}{c}=c^2-\frac{1}{a}=d=2$$

and so on for other $d$.

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III. Cardano's Method

Using Cardano's method of solving cubics, one will have to use the negative square,

$$\gamma=\sqrt{-z^2\,}=\sqrt{-\left(8d^6-16d^3+9+(2d^2-3)\sqrt{4d^3-3\,}\right)^2\,}$$

Thus, for real $d$ such that $\sqrt{4d^3-3}\,$ is real, then both cubics are casus irreducibilis and the sextic will have six real roots as in the above example $d=2$. However, if $\sqrt{4d^3-3}\,$ is complex, then the sextic will have six complex roots like when $d=-2$.

A special case is when $d=1$ as the sextic will have a zero root. However, the other cubic is still usable and quite familiar as,

$$x^3-x^2-2x+1=0$$

the roots of which, correctly ordered are $(a,b,c)=(-2\cos \frac{2\pi}7,\, -2\cos \frac{6\pi}7,\, -2\cos \frac{4\pi}7)$ implying the nice trigonometric identities,

$$a^2-\frac{1}{b}=b^2-\frac{1}{c}=c^2-\frac{1}{a}=1$$ $$abc(a+b+c)=-1$$

Answer 3

Given

$$a^2-b^{-1}=b^2-c^{-1}=c^2-a^{-1}=d. $$

So we have

$$b a^2 - b d - 1=c b^2 - c d - 1=a c^2 - a d - 1=0. $$

Using Groebner basis computations, we have

$$x(b a^2 - b d - 1)+y(c b^2 - c d - 1)+z(a c^2 - a d - 1)\\ =(a-b) \left(a^2 b c+a b^2 c+a b c^2+d^2\right)=0. $$

By condition $a-b\ne 0$, then $a^2 b c+a b^2 c+a b c^2+d^2=0$, which is equivalent to

$$abc(a+b+c)=-d^2. $$

Here my $x,y,z$ is

$$ x=\\\tiny -\left(\left(c^2-d\right) \left(a^3 c (b d-1) \left(c^2-d\right)+a^2 c \left((b d-1) \left(c^2-d\right) \left(c^2 d+c-d^2\right)-b d\right)+a d \left(-\left(b^3-2\right) c^3+\left(b^3-2\right) c d-c^4 \left(b^2+2 b d^2-3 d\right)+c^6 (b d-1)+c^2 d \left(b \left(b+d^2\right)-2 d\right)-b d\right)-c^2 d (b d-1) \left(c^2-d\right)+1\right)\right), \\ y=\\\tiny a^3 \left(c^2-d\right) \left(c^4 d-c^3-2 c^2 d^2+d^3\right)+a^2 \left(-c^6+\left(6 c^3-1\right) c d^4-4 c^2 d^5+\left(-4 c^6+3 c^3-1\right) d^3+\left(c^6+c^3+1\right) c d+\left(c^6-3 c^3+1\right) c^2 d^2+d^6\right)+a \left(b^2 d \left(d \left(c^2-d\right)^2-1\right)+b (c d+1) \left(d \left(c^2-d\right)^2-1\right)+d \left(c^2-d\right) \left(\left(3 c^3-1\right) d^3+c^3+\left(c^3-1\right) c^4 d+\left(2-3 c^3\right) c^2 d^2-c d^4\right)\right)+c \left(c^6 \left(-d^2\right)+3 c^4 d^3-c^2 \left(3 d^4+d\right)+c+d^5\right),\\ z=\\\tiny a^4 b c (b d-1) \left(c^2-d\right)+a^3 b c \left((b d-1) \left(c^2-d\right) \left(c^2 d+c-d^2\right)-1\right)+a^2 \left(-c d \left(b c (b+c) \left(b^2 c+c^3-1\right)+1\right)-d^3 \left(b^2 \left(2 c^4+c\right)+2 b c^2+2 c^3-1\right)+c d^4 \left(b^2 c+1\right)+d^2 \left(b^4 c+b^3 c^2+b^2 \left(c^6+2 c^3-1\right)+b c \left(3 c^3-1\right)+c^2 \left(c^3-2\right)\right)+b c^2 \left(b c^2-1\right)\right)+a \left(-b^4 c d+b^3 \left(-c^2\right) d+b^2 \left(2 c^5 d^3+c^5-c^3 d^4+c^2+\left(-c^6+c^3+1\right) c d^2-\left(c^6+c^3-1\right) d\right)+b (c d+1)^2+d \left(c^3-c d-1\right) \left(c^2 d+c-d^2\right)^2\right)-b^2 c^3 \left(c d \left(d \left(c^2-d\right)^2+1\right)-1\right)+b c^2 d+d \left(c \left(c^5+\left(3 c^3-1\right) d^4+\left(c^3-1\right) c^3 d+\left(c^6-3 c^3+1\right) c d^2-3 \left(c^3-1\right) c^2 d^3-c d^5\right)-1\right). $$

Note that the $x,y,z$ above are given by Mathematica, so it is too difficult to check the proof manually. There should be shorter forms for $x,y,z$ that satisfy the equality. Attention is all you need. :)

btw: You can run the following code in Mathematica to check.

(*Output: (a-b) (a^2 b c+a b^2 c+a b c^2+d^2)*)
bas = {b  a^2 - b  d - 1, c  b^2 - c  d - 1, a  c^2 - a  d - 1};
s = {-((c^2 - d)  (1 - a^3  c^3 - a^2  c^4 + a^3  c  d - 
    a^2  b  c  d + a^2  c^2  d + 2  a  c^3  d + a^3  b  c^3  d - 
    a  b^3  c^3  d + c^4  d + a^2  b  c^4  d - a  b^2  c^4  d - 
    a^2  c^5  d - a  c^6  d - a  b  d^2 - 2  a  c  d^2 - 
    a^3  b  c  d^2 + a  b^3  c  d^2 - c^2  d^2 - 
    a^2  b  c^2  d^2 + a  b^2  c^2  d^2 + 2  a^2  c^3  d^2 + 
    3  a  c^4  d^2 - b  c^4  d^2 + a^2  b  c^5  d^2 + 
    a  b  c^6  d^2 - a^2  c  d^3 - 2  a  c^2  d^3 + b  c^2  d^3 - 
    2  a^2  b  c^3  d^3 - 2  a  b  c^4  d^3 + a^2  b  c  d^4 + 
    a  b  c^2  d^4)), 
    -a  b + c^2 - a^3  c^5 - a^2  c^6 - 
    a  b^2  d + a^2  c  d - a  b  c  d - c^3  d + a^3  c^3  d + 
    a^2  c^4  d + a  b  c^4  d + a  c^5  d + a^3  c^6  d + 
    a^2  c^7  d + a^2  c^2  d^2 - 2  a  b  c^2  d^2 - a  c^3  d^2 - 
    3  a^3  c^4  d^2 + a  b^2  c^4  d^2 - 3  a^2  c^5  d^2 + 
    a  b  c^5  d^2 - a  c^6  d^2 - c^7  d^2 + a^2  c^8  d^2 + 
    a  c^9  d^2 - a^2  d^3 + a  b  d^3 + 3  a^3  c^2  d^3 - 
    2  a  b^2  c^2  d^3 + 3  a^2  c^3  d^3 - 2  a  b  c^3  d^3 + 
    3  a  c^4  d^3 + 3  c^5  d^3 - 4  a^2  c^6  d^3 - 4  a  c^7  d^3 -
     a^3  d^4 + a  b^2  d^4 - a^2  c  d^4 + a  b  c  d^4 - 
    3  a  c^2  d^4 - 3  c^3  d^4 + 6  a^2  c^4  d^4 + 6  a  c^5  d^4 +
     a  d^5 + c  d^5 - 4  a^2  c^2  d^5 - 4  a  c^3  d^5 + a^2  d^6 + 
    a  c  d^6, 
   a  b - a^3  b  c - a^2  b  c^2 + a  b^2  c^2 - a^4  b  c^3 + 
    b^2  c^3 - a^3  b  c^4 + a^2  b^2  c^4 + a  b^2  c^5 - d + 
    a  b^2  d - a^2  c  d + 2  a  b  c  d + a^4  b  c  d - 
    a  b^4  c  d - a  c^2  d + b  c^2  d + a^3  b  c^2  d + 
    a^2  b^2  c^2  d - a  b^3  c^2  d + a^2  b  c^3  d - 
    a  b^2  c^3  d + a^4  b^2  c^3  d - a^2  b^4  c^3  d - 
    b^2  c^4  d + a^3  b^2  c^4  d - a^2  b^3  c^4  d + a  c^5  d - 
    a^3  b  c^5  d - a^2  b^2  c^5  d + c^6  d - a^2  b  c^6  d - 
    a  b^2  c^6  d - a^2  b^2  d^2 - a^2  b  c  d^2 + a  b^2  c  d^2 -
     a^4  b^2  c  d^2 + a^2  b^4  c  d^2 - 2  a^2  c^2  d^2 + 
    a  b  c^2  d^2 - a^3  b^2  c^2  d^2 + a^2  b^3  c^2  d^2 - 
    3  a  c^3  d^2 + 2  a^3  b  c^3  d^2 + 2  a^2  b^2  c^3  d^2 - 
    c^4  d^2 + 3  a^2  b  c^4  d^2 + a  b^2  c^4  d^2 + 
    a^2  c^5  d^2 + a^3  b^2  c^5  d^2 + 2  a  c^6  d^2 + 
    a^2  b^2  c^6  d^2 + c^7  d^2 - a  b^2  c^7  d^2 - b^2  c^8  d^2 +
     a^2  d^3 + 2  a  c  d^3 - a^3  b  c  d^3 - a^2  b^2  c  d^3 + 
    c^2  d^3 - 2  a^2  b  c^2  d^3 - 2  a^2  c^3  d^3 - 
    2  a^3  b^2  c^3  d^3 - 5  a  c^4  d^3 - 2  a^2  b^2  c^4  d^3 - 
    3  c^5  d^3 + 2  a  b^2  c^5  d^3 + 2  b^2  c^6  d^3 + 
    a  c^7  d^3 + c^8  d^3 + a^2  c  d^4 + a^3  b^2  c  d^4 + 
    4  a  c^2  d^4 + a^2  b^2  c^2  d^4 + 3  c^3  d^4 - 
    a  b^2  c^3  d^4 - b^2  c^4  d^4 - 3  a  c^5  d^4 - 3  c^6  d^4 - 
    a  d^5 - c  d^5 + 3  a  c^3  d^5 + 3  c^4  d^5 - a  c  d^6 - 
    c^2  d^6};
Factor[bas . s]

Answer 4

A way to solve it with the help of a symbolic processor and Groebner basis

solrd = Eliminate[{Numerator[Together[a^2 - 1/b - d]] == 0, 
                   Numerator[Together[b^2 - 1/c - d]] == 0, 
                   Numerator[Together[c^2 - 1/a - d]] == 0, 
                   (a + b + c) a b c ==  r}, {a, b, c}] // FullSimplify
(* (d^2 + r) (-27 + 9 d (-4 + d^3) r - 6 d^2 r^2 + r^3) == 0 *)
Solve[solrd, r]