# Solve an equation

#### anemone

##### MHB POTW Director
Staff member
Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.

#### jacks

##### Well-known member
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 $\boxed{y = k - k^3}$

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#### anemone

##### MHB POTW Director
Staff member
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

##### Well-known member
Hi, anemone and jacks!

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

##### MHB POTW Director
Staff member
Hi, anemone and jacks!

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?

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

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

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

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

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