Determine the value of $(a+b+c)^3$ from system of equations

Question

Given the equations

\begin{align} \frac{1}{ab} &=b+2c \\ \frac{1}{bc} &= 2c+3a \\ \frac{1}{ca} &=3a+b,\end{align}

determine the value of $(a+b+c)^3.$

I tried $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=2b+6a+4b$

And then I tried to connect this with $3=ab^2+2bc^2+3ca^2+6abc$ but then I didn't know what to do.

Answer 1

We have $$\frac{\frac{1}{ab}}{\frac{1}{bc}}=\frac{b+2c}{2c+3a}$$ and $$\frac{\frac{1}{ab}}{\frac{1}{ca}}=\frac{b+2c}{3a+b}$$ or $$2c^2=a(b-c)$$ and $$b^2=3ac-bc.$$ Thus, $$a=\frac{2c^2}{b-c},$$ which gives $$b^2=\frac{6c^3}{b-c}-bc$$ or $$b^3-bc^2-6c^3=0$$ or $$b^3-2b^2c+2b^2c-4bc^2+3bc^2-6c^3=0$$ or $$(b-2c)(b^2+2bc+3c^2)=0$$ or $$b=2c,$$ which gives $$a=2c.$$ Can you end it now?

I got the following solution of the system. $$\left\{\left(\frac{1}{\sqrt[3]2}, \frac{1}{\sqrt[3]2},\frac{1}{2\sqrt[3]2}\right)\right\}$$

Answer 2

Here is a way to obtain $a,b,c$ using linear algebra :

$$\begin{cases} b+2c&=&\frac{1}{ab}\\ 2c+3a&=&\frac{1}{bc} \\ 3a+b&=&\frac{1}{ca}\end{cases}\tag{1}$$

can be expressed equivalently under the form :

$$\begin{pmatrix}0&1&2\\3&0&2\\3&1&0\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\dfrac{1}{abc}\underbrace{\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}_C\begin{pmatrix}a\\b\\c\end{pmatrix}$$

($C$ is a permutation matrix). Multiplying LHS and RHS by $C^{-1}$ :

$$\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\begin{pmatrix}0&1&2\\3&0&2\\3&1&0\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\dfrac{1}{abc}\begin{pmatrix}a\\b\\c\end{pmatrix}$$

$$\underbrace{\begin{pmatrix}3&0&2\\3&1&0\\0&1&2\end{pmatrix}}_N\underbrace{\begin{pmatrix}a\\b\\c\end{pmatrix}}_V=\underbrace{\dfrac{1}{abc}}_{\lambda}\underbrace{\begin{pmatrix}a\\b\\c\end{pmatrix}}_V$$

Therefore $V$ is an eigenvector of $N$ associated with eigenvalue $\lambda$.

As matrix $N$ has eigenvalues $4, 1+\sqrt{2}i, 1-\sqrt{2}i$, and as we are looking for a real (!) candidate, we can conclude that

$$\dfrac{1}{abc}=4\tag{2}$$

(2) allows to transform system (1) into the homogeneous linear system :

$$\begin{cases} b+2c&=&4c\\ 2c+3a&=&4a \\ 3a+b&=&4b\end{cases}\tag{3}$$

whose almost immediate solution is

$$a=k2, b=k2, c=k1,\tag{4}$$

a result that we could have as well obtained using the information coming from the eigenvector associated with $\lambda$, which is

$$\begin{pmatrix}a\\b\\c\end{pmatrix}=k\begin{pmatrix}2\\2\\1\end{pmatrix}$$

(one can check it easily).

Therefore, plugging (3) into system (1) gives the value of $k$, yielding the values of $a,b,c$ as in the solution by Michael Rozenberg.

Remarks :

1) I don't pretend that this solution is short, but it has the advantage of being a systematic method that one can think about in certain "a priori" nonlinear systems of such a type.

2) It must be said that system (3) is nothing else that the system one has to solve for obtaining the eigenvector associated with eigenvalue $\lambda=4$.