Solve equation: $y+k^3=\sqrt[3]{k-y}$

[anemone]

Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.

 

[juantheron]

Spoiler

Consider function $f(k) = y + k^3$, then $f^{-1}(k) = \sqrt[3]{k - y}$. Hence \[f^{-1}(k) = f(k)\] This can happen if and only if \[k = f(k) = f^{-1}(k)\] i.e. \[k = \sqrt[3]{k - y} = y + k^3\] So \[\boxed{y = k - k^3}\]

 

[anemone]

Spoiler

Consider function $f(k) = y + k^3$, then $f^{-1}(k) = \sqrt[3]{k - y}$. Hence \[f^{-1}(k) = f(k)\] This can happen if and only if \[k = f(k) = f^{-1}(k)\] i.e. \[k = \sqrt[3]{k - y} = y + k^3\] So \[\boxed{y = k - k^3}\]

Hey jacks, thanks for participating and your solution is simple, elegant and nice! Well done, jacks!:)

 

[lfdahl]

Hi, anemone and jacks!

Thankyou for your answers. Jacks solution is very elegant!(Star)

I have one question. I do not understand why the following implication is true:
$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $$

Why is k appearing on the RHS?

I would deduce the following:
$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$
By definition:
$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$

 

[anemone]

Hi, anemone and jacks!

Thankyou for your answers. Jacks solution is very elegant!(Star)

I have one question. I do not understand why the following implication is true:
$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $$

Why is k appearing on the RHS?

I would deduce the following:
$$ f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$
By definition:
$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$

Hi lfdahl,

I am sorry for I only replied to you days after...I thought to myself to let jacks to handle it and I would only chime in if we didn't hear from jacks 24 hours later. But it somehow just slipped my mind.:(

Back to what you asked us...I believe if we use the identity

$f(f^{-1}(k))=k$,

and that for we have $f(k)=y+k^3$, we would end up with getting $f^{-1}(k)=\sqrt[3]{\mathbf{k}-y} $, does that answer your question, lfdahl

Spoiler

$f(k)=y+k^3$

$f(f^{-1}(k))=k$

$y+(f^{-1}(k))^3=k$

$(f^{-1}(k))^3=k-y$

$f^{-1}(k)=\sqrt[3]{k-y}$