Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$ Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)
I attempted to solve this question as follows:
$32y+32x^3=6x+17$
$y=-x^3+\frac{6x}{32}+\frac{17}{32}$
$y-\frac{1}{2}=-x^3+\frac{3x}{16}+\frac{17}{32}-\frac{1}{2}$
Here it started getting very complex, and hence I don't think it can be solved this way. After this I tried doing something similar with the other two equations, but once again it was ending up way too complex. It is obvious that the solution is $x=y=z=\frac{1}{2}$, but I can't manage to prove it. Could you please explain to me how to solve this question?

Just for simplicity's sake, make a change of variables $$(a,b,c) = (2x,2y,2z)$$
We have the system of equations
\begin{align}
16b + 4a^3 &= 3a+17 \\
8c + 4b^3 &= 3b+9 \\
4a + 4c^3 &= 3c+5 \\
\end{align}
or
\begin{align}
16(b-1)  &= - 4a^3+3a+1 \\
8(c-1)   &= - 4b^3+3b+1 \\
4(a-1)  &=-4c^3+ 3c+1 \\
\end{align}
or
\begin{align}
16(b-1)  &= - (a-1)(4a^2+4a+1) \tag{1}\\
8(c-1)   &= - (b-1)(4b^2+4b+1) \tag{2}\\
4(a-1)  &=-(c-1)(4c^2+4c+1) \tag{3}\\
\end{align}
Suppose $b >1$ then from $(1)$: $a<1$ $\implies$ from $(3)$: $c>1$ $\implies$ from $(2)$: $b>1$ $\implies$ contracdiction!
Same for  $b <1$, we have also a contradiction.
For $b=1$, we have $c =1$ and $a =1$.
Conclusion: the system of equations has a unique solution $(a,b,c) = (1,1,1)$ or $$(x,y,z) = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2} \right)$$
