<s>[INST] If $f(x+y^3)=f(x)+[f(y)]^3$ and $f'(0)\ge0$, what can $f(10)$ be? A real valued function satisfies the condition: $f(x+y^3)=f(x)+[f(y)]^3$ for all real $x$, $y$.
If $f'(0)\ge0$ how to find $f(10)$ ?
 [/INST] Here is my attempt:
Let $(x,y)=(0,t)$:$$
f(0+t^3) = f(t^3) = f(0) + [f(t)]^3
$$We can use this to find $f(0)$:$$
f(0^3)=f(0)+[f(0)]^3 \implies f(0) - f(0) = [f(0)]^3 \implies f(0) = 0
$$This means our function simplifies to:$$
f(t^3) = [f(t)]^3
$$Next let $(x,y)=(t,c)$ where $c$ is some arbitrary constant with respect to $t$:$$
f(t+c^3) = f(t) + [f(c)]^3
$$Differentiate both sides with respect to $t$:$$
f'(t+c^3) = f'(t) \implies f'(c^3) = f'(0) \geq 0
$$Since $c$ is any arbitrary constant, this means that $f'$ is always non-negative and so $f$ is always increasing or constant. So $0<1 \implies f(0)\leq f(1) \implies 0 \leq f(1)$.
Next let $(x,y)=(0,1)$:$$
f(1^3)=f(1) = [f(1)]^3 \implies f(1) = 1 \lor f(1) = 0
$$The fact that $0 \leq f(1)$ rules out $f(1)=-1$.
Next let $(x,y)=(1,1)$:$$
f(1+1^3) = f(1) + [f(1)]^3 \implies f(2) = 1 + 1^3 = 2 \lor f(2) = 0
$$Next let $(x,y)=(2,2)$:$$
f(2+2^3) = f(2) + [f(2)]^3 \implies f(10) = 2 + 2^3 = 10 \lor f(10) = 0
$$

It would appear that $f(x)=x$ satisfies both conditions of the provided function.$$
f'(0) = 1 \geq 0 \quad \land\quad f(x+y^3) = x+y^3 = f(x) + [f(y)]^3
$$
It would appear that $f(x)=0$ also satisfies both conditions.$$
f'(0) = 0 \geq 0 \quad \land\quad f(x+y^3) = 0 = 0 + [0]^3
$$
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