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Find all the real solutions to:

$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$

Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?

Thank you.

$$ \begin{align} x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\ x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\ x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\ \sqrt[3]{6+x} &= x \\ x^3 &= 6+x \\ x^3-2x^2+2x^2-4x+3x-6 &= 0 \\ (x-2)(x^2+2x+3) &= 0 \\ x &= 2 \end{align} $$

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  • $\begingroup$ This doesn't answer your question per se since it seems like an ansatz out of left field but.... if $\sqrt[3]{6+x}=x$ then line 2 is true. We'd get $\sqrt[3]{6+\sqrt[3]{6+x}} = \sqrt[3]{6+x}$ $\endgroup$ – Kitter Catter Dec 18 '19 at 15:53
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    $\begingroup$ Consider the function $f \colon x \mapsto \sqrt[3]{6 + x}$. It is continuous, strictly increasing, and has a unique fixed point. The second line looks for a fixed point of $f\circ f\circ f$. The properties of $f$ guarantee that such a point must be the fixed point of $f$. $\endgroup$ – Daniel Fischer Dec 18 '19 at 15:59
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    $\begingroup$ @Elementary No, it works for any finite number of iterations. Let $x_0$ be the fixed point. Then we either have $f(x) > x$ for all $x > x_0$, or $x_0 < f(x) < x$ for all $x > x_0$. In the first case $f^3(x) > f^2(x) > f(x) > x$, and in the second $x_0 < f^3(x) < f^2(x) < f(x) < x$, when $x > x_0$. A similar argument for $x < 0$ shows that $x_0$ is the only fixed point of $f^3$. And it's clear that it also works for $f^n$, where $n$ is an arbitrary strictly positive integer. $\endgroup$ – Daniel Fischer Dec 18 '19 at 16:14
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    $\begingroup$ The only correct answer so far is the comment of @DanielFischer. $\endgroup$ – WhatsUp Dec 18 '19 at 16:31
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    $\begingroup$ If we look at complex solutions, they indeed aren't, but since things are explicitly constrained to be real, things are easier. (Not that that's too usual, often things are much easier in the complex world. But sometimes they are easier in $\mathbb{R}$.) $\endgroup$ – Daniel Fischer Dec 18 '19 at 20:50
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Goal: Establish the fact that in order to solve $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}},$$ we only need to solve $$ x=\sqrt[3]{6+x}. $$

Define $f(x)$:

$$ f(x) = \sqrt[3]{x+6}.$$

From line 3, we are looking to solve $$ x = f(f(f(x))).$$

Notice that $f(x)$ is strictly increasing. In other words, for all $a$ and $b$ in the domain of $f(x)$, if $a > b$, then $f(a) > f(b)$.

This means that the only way in which we can get solutions for $x = f(f(f(x)))$ is if $x = f(x)$. To see why, assume that $x > f(x)$: $$ x > f(x) \Rightarrow \\ f(x) > f(f(x)) \Rightarrow \\ f(f(x)) > f(f(f(x))) \Rightarrow \\ x > f(x) > f(f(x)) > f(f(f(x))). $$ The statement $x = f(f(f(x)))$ is now untrue. The exact same reasoning can be used for the assumption that $x < f(x)$.

Since $x < f(x)$ and $x > f(x)$ do not give us any solutions, then $x = f(x)$ is the only case we have left.

FYI, the values of a function's domain that don't change when passed into the function are called the "fixed points" of $f(x)$. In other words, all $x$ such that $x = f(x)$ are called the fixed points of $f(x)$.

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I think the step that is is missing is

$$x = \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{\color{red}x+6}}}$$

$$=\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}x}}}$$

$$= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}{\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{blue}x}}} }}}}, $$

and we can continue to obtain the infinite expression

$$x = \sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$

(This is actually invalid and handwavey and not legitimate... but let's go with it.)

Then, we can do

$$x = \sqrt[3]{6 + \color{blue}{\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}}}=\sqrt[3]{6 + \color{blue}x}.$$

And then, we continue.

The problem is, of course, what the #@%! is $\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$ supposed to mean? Is that even well defined and can we do math on it?

And the answer is, yes.

If $a_0=6$ and $a_k = \sqrt[3]{6 + a_{k-1}}$ for $k>1$, then we can prove by induction that $0 < a_{k+1} < a_k\le 6$, with equality holding if and only if $k =0$. Therefore, $\{a_k\}$ is monotonically decreasing and bounded below, so $\lim_{n\to \infty} a_n = x$ for some real $x$, which if we wanted to, we could express as $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$ if we took $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}:= \lim_{n\to \infty} a_n$$ as a definition.

Now, for converging limits, $$x = \lim_{n\to \infty} a_n = \lim_{n\to \infty}a_{n+1} = \lim_{n\to\infty}\left(\sqrt[3]{6+a_n}\right) = \sqrt[3]{6 + \lim_{n\to \infty}a_n} = \sqrt[3]{6 + x}.$$

So we can do it.


I guess we can also do it without resorting to the infinite.

Let $?$ be notation for $<$ or for $>$ or for $+$, where we do not know which. However, since each of $<,>,=$ are transitive and preserved via "adding to both sides" and "cubing both sides", we may do the following manipulations.

If $x \enspace?\enspace \sqrt[3]{6+x}$, then

$$6+ x \enspace?\enspace 6 +\sqrt[3]{6+x}\Rightarrow$$

$$\sqrt[3]{6 + x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$

$$x\enspace?\enspace\sqrt[3]{6 + x}\enspace?\enspace\sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$

$$\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}\Rightarrow$$

$$x\enspace?\enspace\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$

But we showed $$x = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$

So by transitivity, $$x \enspace?\enspace x.$$ But by trichotomy, $x =x$ and $x \not < x$ and $x \not > x$. So it must be that $?$ is notation for $=$, and $x = \sqrt[6]{x+6}$.

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Begin as in OP $\begin{align} x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \tag{1}\\ x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \tag{2}\\ x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \tag{3} \end{align}$

Setting $t=\sqrt[3]{6+x}$, the equation $(3)$ rewrites $$t^3-6=\sqrt[3]{6+\sqrt[3]{6+t}},$$ equivalent to $(1).$ Hence $t=x.$

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So the equation is as follows $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$ Notice that when you repetedly put back $x$ in the RHS, you get the following infinite sum. $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+{\cdots}}}}}}}$$ From here, you can replace the part from the first cube root with $x$ to get $$x=\sqrt[3]{6+x}$$

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    $\begingroup$ This is not a rigorous answer. Talking about infinitely nested roots without mentioning convergence is logically wrong. $\endgroup$ – WhatsUp Dec 18 '19 at 16:13

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