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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.

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

Steven Stadnicki
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  • 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 $\sqrt[3]{6+\sqrt[3]{6+x}} = \sqrt[3]{6+x}$ – Kitter Catter Dec 18 '19 at 15:53
  • The idea is to see this as a sequence of iterations $x_0 = x$ and $x_{n+1} = \sqrt[3]{6+x_n}$, so your result is a fixed point of this iteration. But that does not guarantee that this is the only possible solution only after a finite amount of steps... – gt6989b Dec 18 '19 at 15:58
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    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 $f$. – Daniel Fischer Dec 18 '19 at 15:59
  • @DanielFischer I thought we could only use this method when there were an infinite number of steps. –  Dec 18 '19 at 16:03
  • @DanielFischer Well done, make an answer out of your comment and get reputation points! ;) – Olivier Roche Dec 18 '19 at 16:05
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    @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. – Daniel Fischer Dec 18 '19 at 16:14
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    The only correct answer so far is the comment of @DanielFischer. – WhatsUp Dec 18 '19 at 16:31
  • @DanielFischer Actually $f(x) < x$ for $x > 2$ and $f(x) > x$ for $x < 2$. – Zarrax Dec 18 '19 at 16:47
  • @Zarrax Yes, for this particular $f$ that is the case (easy to see). But for the general case ($f$ continuous, strictly increasing, with a unique fixed point) there are more possibilities. – Daniel Fischer Dec 18 '19 at 16:51
  • @DanielFischer Given the complex roots, these equations (original and the last) are not equivalent. Do I think right? –  Dec 18 '19 at 20:43
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    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 $\mathbb{R}$.) – Daniel Fischer Dec 18 '19 at 20:50
  • @DanielFischer It is very curious for me Only the real roots are equal of these equations. In other words $\text{Real roots of} (x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6)=\text{Real roots of} (\sqrt[3]{x+6}=x)$ but $\text{Complex roots of} (x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6) \neq \text{Complex roots of} (\sqrt[3]{x+6}=x)$ the original equation is a polynomial with a degree of $24$. Therefore, our method works to find only the real roots. I hope I'm getting the right result. –  Dec 18 '19 at 21:09

4 Answers4

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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 $f(x)$.

zhuli
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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 $x = \sqrt[6]{x+6}$.

ViHdzP
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fleablood
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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 $t=x.$

EDIT

The functions $f:x\to x$ and $g:x\to \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}}\;$ are

  • continuous on $[0,\infty)$
  • increasing ($f$ is increasing trivially, $g$ is composed from increasing functions)
  • $f$ is linear, $g$ is concave
  • $f(0)=0<g(0), f(21)=21>g(21)$

Hence there exists unique solution to $(3).$ If $t$ and $x$ both satisfy, then they are interchangeable.

Viera Čerňanová
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  • This logic is not complete. Take for instance $x^2 = 1$ and $t^2 = 1$. Just because the equations are equivalent does not imply that $x=t$. You could have $x=1$ and $t=-1$. – zhuli Jun 09 '21 at 19:59
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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 $$x=\sqrt[3]{6+x}$$

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