Find all the $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f\left(f^3(x)+y^3\right)=x^2+f^3(y)$.

Question

Find all the $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f\left(f^3(x)+y^3\right)=x^2+f^3(y)$, where $f^3(x)$ stands for $[f(x)]^3$.

I really don't know where to start off.

Answer 1

Let $P(x, y)$ represent the given equation

$$f([f(x)]^3+y^3)=x^2+[f(y)]^3$$

$P(x, f(y))$:

$$f([f(x)]^3+[f(y)]^3)=x^2+[f(f(y))]^3$$

Now exploit symmetry by switching $x, y$:

$$f([f(y))]^3+[f(x)]^3)=y^2+[f(f(x))]^3$$

Comparing gives $$y^2+[f(f(x))]^3=x^2+[f(f(y))]^3$$

Put $y=1$ to get $[f(f(x))]^3=x^2-1+[f(f(1))]^3=x^2+c$ for some constant $c$.

$P(x, 0)$:

$$f([f(x)]^3)=x^2+[f(0)]^3$$

$P(0, y)$:

$$f([f(0)]^3+y^3)=[f(y)]^3$$

Apply $f$ to both sides to get

$$f(f([f(0)]^3+y^3))=f([f(y)]^3)$$

$$(([f(0)]^3+y^3)^2+c)^{\frac{1}{3}}=y^2+f(0)^3$$

$$([f(0)]^3+y^3)^2+c=(y^2+[f(0)]^3)^3$$

$$y^6+2[f(0)]^3y^3+([f(0)]^6+c)=y^6+3[f(0)]^3y^4+3[f(0)]^6y^2+[f(0)]^9$$

Since this holds for all $y$, we must have $$2[f(0)]^3=0, 3[f(0)]^3=0, 3[f(0)]^6=0, [f(0)]^6+c=[f(0)]^9$$ Thus $f(0)=c=0$.

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Let us pause for a moment and take stock of what we have:

$$f(0)=0, c=0$$

$$f(f(x))=(x^2+c)^{\frac{1}{3}}=x^{\frac{2}{3}}$$

$$f([f(x)]^3)=x^2+[f(0)]^3=x^2$$

$$[f(y)]^3=f([f(0)]^3+y^3)=f(y^3)$$

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Now substituting $y=1$ into the fourth equation above gives $f(1)^3=f(1)$ so $f(1)=0, 1$. However if $f(1)=0$, then $f(f(1))=f(0)=0 \not =1^{\frac{2}{3}}$, a contradiction. Thus $f(1)=1$.

$P(1, 1)$:

$$f([f(1)]^3+1)=1+[f(1)]^3$$

Since $f(1)=1$, we get $f(2)=2$. Then $2^{\frac{2}{3}}=f(f(2))=f(2)=2$, a contradiction.

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Therefore no such functions satisfying the given equation exist.