Solving the equation $ f^{-1}(x)=f(x)$ I attempted to solve the equation given in the title for the function;
$$f: \mathbb R_{++} \to\mathbb R_{++}; \quad f(x)=x^2(x+2)$$
I understand that the problem is equivalent to solving $f(f(x))=x$
but since this seemed like too much work, I had a look at the solution and it stated that;
$$ f^{-1}(x)=f(x) \Longrightarrow  f(x)=x$$
I don't understand why this is the case. Can someone please explain this?
Thanks 
 A: Although $f(f(x))=x\implies f(x)=x$ is not always true, it is true if $f$ is an increasing function (even if only weakly increasing).  
Suppose that $f(x)=y$ and $f(y)=x$. If $x\le y$, then $f(x)=y\le f(y)=x$, and vice-versa.
So either way, both $x\le y$ and $y\le x$ are true, and thus $x=y$ and $f(x)=x$.
So your book is not off the rails; this is a valid approach to the problem.
A: Since $f: \mathbb R_{++} \to \mathbb R_{++}$ is settled in the comments, your idea is good. Write $f(x) = x^3 + 2x^2$ to reduce the complexity of $f(f(x))$ and then solve
$$f(f(x)) = (x^3+2x^2)^3 + 2(x^3 + 2x^2)^2 \stackrel!= x \qquad x>0$$
We can expand this to
$$x^9 + 6x^8 + 12x^7 + 8x^6 + 2x^6 + 8x^5 + 8x^4 - x = 0$$
Then divide by $x$ since $x\ne 0$ for
$$x^8 + 6x^7 + 12x^6 + 8x^5 + 2x^5 + 8x^4 + 8x^3 - 1 = 0$$
WolframAlpha tells us there is only one positive root at
$$x = \sqrt 2 - 1$$
A: If the problem were to find a solution of $f^{-1}(x)=f(x)$ with $x\gt0$, then the hint would be useful if the implication were reversed, i.e.,

$$f(x)=x\implies f^{-1}(x)=f(x)$$

Setting $x^2(x+2)=x$ leads to the equation
$$x^2+2x-1=0$$
which has positive solution $x=\sqrt2-1$.  
But we can say more.  Borrowing the degree-$8$ polynomial from AlexR's answer, we know that $x^2+2x-1$ must be a factor of it.  And indeed it is:
$$x^8+6x^7+12x^6+10x^5+8x^4+8x^3-1=(x^2+2x-1)(x^6+4x^5+5x^4+4x^3+5x^2+2x+1)$$
and since the coefficients of $x^6+4x^5+5x^4+4x^3+5x^2+2x+1$ are all positive, the degree-$8$ polynomial has no other solutions with $x\gt0$.  So $x=\sqrt2-1$ is not only a solution, it's the only solution.
A: Answering this problem is simple.
f(x) = x^2(x+2)
Thus f(x) = a(x)b(x) where a(x) = x^2 and b(x) = x+2
The product rule tells us that:
f'(x) = 3x^2 + 4x
Thus solve for f(x) = f(x)':
x^2(x+2) = 3x^2 + 4x
x(x^2 -x - 4) = 0
Therefore x = 0; x = (1±sqrt[17])/2
