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

Question

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.

Answer 1

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 $$x=a-a^3$$

Answer 2

Instead of cuberooting, cube.

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

Answer 3

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. :-)

Answer 4

Making $a-x = y^3$

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

adding both equations

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