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If $a\in \mathbb{R}$, then show $$x + a^3 = \sqrt[3]{a-x}$$

I am a newbie to such questions, that seem more like transformation of one form to another. But, can take a more elegant form by using the recursiveness (could not find a better word) as follows:

$a = \sqrt[3]{\sqrt[3]{a-x}-x}$

Also, can define the function :
$f(f(a))=f(\sqrt[3]{a-x}) = \sqrt[3]{\sqrt[3]{a-x}-x}$

Now, need form a relation between $f(a)$ & $a$. But, how is not clear.

nonuser
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jiten
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    What happens when $a=0$? – Topology Jun 01 '18 at 18:51
  • @Topology Do you mean that at $a=0$ there is no mapping possible by nested function. If so, please elaborate. – jiten Jun 01 '18 at 18:53
  • Do you need to solve the equation or just show that it has solutions? – Arnaud Mortier Jun 01 '18 at 18:55
  • @ArnaudMortier Both, but showing that solution exists is taken by me as a preliminary step. – jiten Jun 01 '18 at 18:55
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    Showing that there is one unique solution is easy since one function is bijective increasing and the other bijective decreasing. But in fact you have a great answer by @ChristianF below. – Arnaud Mortier Jun 01 '18 at 18:58
  • @ArnaudMortier I could not gather which is bijective increasing, & which one is bijective decreasing. May be your intent is shown up in the answer stated by you, but not clear to me. Please elaborate. – jiten Jun 01 '18 at 19:00
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    $x+\text{constant}$ is bijective increasing, $\sqrt[3]{\text{constant}-x}$ is bijective decreasing. – Arnaud Mortier Jun 01 '18 at 19:01
  • You need to clarify the question, because as written it's nonsense. It's impossible to show $x + a^3 = \sqrt[3]{a-x}$ iif $a\in\Bbb R$, because it simply doesn't follow. For example let $x=a=1$. If the problem is actually to solve that equation then say so! – David C. Ullrich Jun 01 '18 at 19:06
  • @ArnaudMortier By stating that it is easy to show that unique soln. exists, do you mean that the only solution is where the two functions concur. But, how to show by some technique that both functions have a common value. I hope you have shown only the uniqueness of the common solution only. Please place an answer for all this. – jiten Jun 01 '18 at 19:30
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    prove or disprove the following: if $f: \mathbb{R} \to \mathbb{R}$ is a continuous increasing bijection function and $g:\mathbb{R} \to \mathbb{R}$ is a continuous decreasing bijection function. then $h(x)=f(x)-g(x)$ is a bijection function. – Siong Thye Goh Jun 02 '18 at 06:32
  • @SiongThyeGoh Am not good at analysis, & am thankful (as usual) for goading me to improve. Have put a post, as it was coming into 2-3 comments for one response only. It is at : https://math.stackexchange.com/q/2805719/424260. I have got response, but the significance of 'proving' (as per the answer there) to this question & the highest voted answer is not clear at all. The reason is that here the two functions have been proved to have only one unique point in common as common domain. – jiten Jun 02 '18 at 22:31
  • @SiongThyeGoh Please see my post for intuitively seeing that the G.F. for having not more than $k$ objects is given by : $\frac{1-x^{k+1}}{1-x}$. It is at : https://math.stackexchange.com/q/2806009/424260. – jiten Jun 03 '18 at 00:09
  • @SiongThyeGoh My second last comment is in response to your question, so expect vetting of my post. – jiten Jun 03 '18 at 03:28

4 Answers4

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Let $t=\sqrt[3]{a-x}$, then we can rewrite equation like this:

$$a-t^3+a^3 = t\implies (a-t)(a^2+at+t^2) +(a-t)=0 $$ so $t=a$, since $a^2+at+t^2 +1 >0$. Now we have $$x=a-a^3$$

nonuser
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    Very nice... +1 – Arnaud Mortier Jun 01 '18 at 18:56
  • could not gather how $a-t^3 +a^3 =t$. Stumbling too much here. – jiten Jun 01 '18 at 19:09
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    $t^3 =a-x$ so $x=a-t^3$ – nonuser Jun 01 '18 at 19:10
  • You seem to have manipulated to achieve the goal of having the bijective (although cannot prove it, or think of a way for that) monotonically increasing function form. I am stating this in context of @ArnaudMortier comment to the OP. – jiten Jun 01 '18 at 19:17
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    he has simplifed the expression until $(a-t)(a^2+at+t^2+1)=0$, since $a^2+at+t^2+1>0$, he can divide that term on both sides and we are left with $a-t=0$. An alternative view is once you have $a^3+a=t^3+t$, check that $x^3+x$ is an increasing function. – Siong Thye Goh Jun 02 '18 at 06:35
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Making $a-x = y^3$

$$ a-x = y^3\\ x+a^3 = y $$

adding both equations

$$ a+a^3 = y + y^3 \Rightarrow a = y \Rightarrow x = a-a^3 $$

Cesareo
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Instead of cuberooting, cube.

So try: $$(x+a^3)^3=a-x$$ $$x^3+3a^3x^2+3a^6x+a^9=a-x$$ $$x^3+3a^3x^2+(3a^6+1)x+(a^9-a)=0$$

Rhys Hughes
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    Then factor out $x+a^3-a$. – J.G. Jun 01 '18 at 18:58
  • @J.G. It does not seem obvious / trivial to think of factoring out $x+ a^3-a$ that I hope means that a root is $x=a−a^3$. There must be a way to get the root without knowing the answer beforehand. – jiten Jun 02 '18 at 00:20
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    @jiten ChristianF's is the best. The interesting consequence of such a factorisation would be to find the "fictional" roots of the quadratic factor. – J.G. Jun 02 '18 at 06:51
  • In fact, I find the quadratic is $x^2+(2a^3+a)x+a^6+a^4+a^2+1$, making the complex roots $a^3-\frac{a}{2}\pm i\sqrt{\frac{3a^2}{4}+1}$. – J.G. Jun 02 '18 at 07:00
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Substituting $y=\sqrt[3]{a-x}$, and thus $x = a - y^3$, we get

$$a - y^3 + a^3=y$$

Ordering

$$y^3 - y = a^3 - a$$

Thus, $y=a$ is a solution. And, if you have a 3rd degree polynom with a known root, you know what follows. :-)

peterh
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