How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try 
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then 
$$
81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16 =0  
$$
here  $( x \neq 0; x \neq \frac{2}{3}; x \neq - \frac{2}{3}; 27x^3 - 12x + 8 \neq 0 )$.
Is there some simple (and different) way to solve it?
 A: This is really a very unpleasant problem.
If you consider the function $$f(x)=81 x^5-81 x^4-90 x^3+36 x^2+16 x-16$$ its derivative $$f'(x)=405 x^4-324 x^3-270 x^2+72 x+16$$ shows four roots (which are supposed to be expressed with radicals - see here) but they are so complex that I shall not put the solution here. Their approximate values are $$x_1^*\approx -0.6061944$$ $$x_2^*\approx -0.1533643$$ $$x_3^*\approx +0.3518603$$ $$x_4^*\approx +1.2076984$$ You could show (very funny without a calculator !) that, for all $x_i^*$'s, $f(x_i^*)<0$. So, there is only one root which will be greater than $x_4^*$. By inspection, since $f(\frac 32)=-\frac{311}{32}$ and $f(2)=736$, you can start Newton method from the left. For this problem, the iterative scheme will be $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{324 x^5-243 x^4-180 x^3+36 x^2+16}{405 x^4-324 x^3-270 x^2+72 x+16}$$ So, starting at $x_0=\frac 32$, the successive iterates will be $$x_1=1.520533474$$ $$x_2=1.519495986$$ $$x_3=1.519493219$$ which is the solution for ten significant figures.
A: the equation $$81 x^5-81 x^4-90 x^3+36 x^2+16 x-16=0$$ can only be solved by a numerical method, e.g. the Newton method
A: This graph helps to show what's happening. 

It's not an answer but a companion to Claude's, in that it helps illustrate the turning points he identified and shows the fact that there's one solution. Note the scales are different, I had to stretch the Y axis. 
