MHB Solve Equation V: Real X Solutions

[anemone]

Find all real $x$ satisfying $x^9+\dfrac{9x^6}{8}+\dfrac{27x^3}{64}-x+\dfrac{219}{512}=0$.

 

[lfdahl]

Here is my solution:

Spoiler

$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512}=0 \\\\
\Rightarrow (x^3+\frac{3}{8})^3-x+\frac{3}{8}=0\; \; so\; \; x^3+\frac{3}{8}=\sqrt[3]{x-\frac{3}{8}} $$
(Recall jacks elegant solution in the thread: http://mathhelpboards.com/challenge-questions-puzzles-28/solve-equation-9126.html)
$$Let \; \; f(x)=x^3+\frac{3}{8} \; \; then \; \; f^{-1}(x)=\sqrt[3]{x-\frac{3}{8}}$$
$$f(x)=f^{-1}(x)\Leftrightarrow f(x)=x$$
So the 9th degree polynomial reduces to a 3rd degree one: $ x^3-x+\frac{3}{8}=0 $
One of the solutions is: $x=\frac{1}{2}$
Polynomial division gives:
\[x^3-x+\frac{3}{8}=(x-\frac{1}{2})(x^2+\frac{1}{2}x-\frac{3}{4})\]
Thus there are three real solutions:
\[x\in \left \{ \frac{1}{2}, \frac{1}{4}(-1\pm \sqrt{13}) \right \}\]

 

[anemone]

Here is my solution:

Spoiler

$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512}=0 \\\\
\Rightarrow (x^3+\frac{3}{8})^3-x+\frac{3}{8}=0\; \; so\; \; x^3+\frac{3}{8}=\sqrt[3]{x-\frac{3}{8}} $$
(Recall jacks elegant solution in the thread: http://mathhelpboards.com/challenge-questions-puzzles-28/solve-equation-9126.html)
$$Let \; \; f(x)=x^3+\frac{3}{8} \; \; then \; \; f^{-1}(x)=\sqrt[3]{x-\frac{3}{8}}$$
$$f(x)=f^{-1}(x)\Leftrightarrow f(x)=x$$
So the 9th degree polynomial reduces to a 3rd degree one: $ x^3-x+\frac{3}{8}=0 $
One of the solutions is: $x=\frac{1}{2}$
Polynomial division gives:
\[x^3-x+\frac{3}{8}=(x-\frac{1}{2})(x^2+\frac{1}{2}x-\frac{3}{4})\]
Thus there are three real solutions:
\[x\in \left \{ \frac{1}{2}, \frac{1}{4}(-1\pm \sqrt{13}) \right \}\]

Well done, lfdahl! You know, I was kind of wondering initially that I wasn't sure if you would think of that particular thread when you saw this challenge problem! You're one of the great problem solvers at MHB and now, I think honesty is one of your best strengths!