Why should it be $\sqrt[3]{6+x}=x$?

Question

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.

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

Answer 1

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 $f(x)$.

Answer 2

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

Answer 3

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.

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

Answer 4

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.