Is there a unique function $f:\Bbb R\to\Bbb R$ satisfying $f(x)^3+3f(x)^2-x^3+2x+3f(x)=0$?
Prove your claim.
Note: $f(x)^3=f(x)\cdot f(x)\cdot f(x)$.
My attempt:
I rewrote the expression like this:
$$f(x)^3+3f(x)^2+3f(x)=x^3-2x$$
Function $g(x):=x^3-2x$ is surjective and differentiable, so I thought the same holds for the LHS and hence for $f(x)$. I factorized the LHS as $f(x)(f(x)^2+3f(x)+3)$.
I argue $f(x)^2+3f(x)+3>0,\forall x\in\Bbb R$ because the parabola $t^2+3t+3$ doesn't have any real roots and is convex.
Since $g(x)$ on the RHS is surjective and $f(x)^2+3f(x)+3>0$, $f(x)$ has to be surjective, too.
Let $h(x)=f(x)^3+3f(x)^2+3f(x)$.
$\begin{aligned}(h(x))'&=3f(x)^2f'(x)+6f(x)f'(x)+3f'(x)\\&=3f'(x)(f(x)^2+2f(x)+1)\\&=f'(x)\underbrace{(f(x)+1)^2}_{\ge 0}.\end{aligned}$
Next: $(g(x))'=3x^2-2$
$\begin{aligned}3x^2-2&=0\\\iff x&=\pm\sqrt{\frac23}\\\implies g'(x)&<0, x\in\left(-\sqrt{\frac23},\sqrt{\frac23}\right).\end{aligned}$
This means $h'(x)<0$ on the same interval and $h'(x)=f'(x)(f(x)+1)^2<0\iff f'(x)<0$, so $f(x)$ should be decreasing on $\left(-\sqrt{\frac23},\sqrt{\frac23}\right)$ as well, and increasing on $\left(-\infty,-\sqrt{\frac23}\right]\cup\left[\sqrt{\frac23},+\infty\right)$.
Function $g(x)=x^3-2x$, from the beginning, is odd, so $h(x)=f(x)^3+3f(x)^2+3f(x)$ should also be odd. I think $f(x)$ shouldn't be odd, because $3f(x)^2$ would be even (while $f(x)^3$ and $f(x)$ would be odd) preventing $h$ from being odd.
I'm neither sure if the information I have are $100\%$ accurate nor relevant. May I ask for advice on solving this task?
Thank you in advance!
It's $$(f(x)+1)^3=x^3-2x+1$$ or $$f(x)=\sqrt[3]{x^3-2x+1}-1.$$