Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$ 
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$ for all $x,y\in\mathbb{R}$. If $f'(0)\ge0$, find $f(10)$.

If $x=y=0$, $f(0+0)=f(0)+f(0)$. So $f(0)=0$.
If $x=0$, $f(0+y^3)=f(0)+f(y)^3$. So $f(y^3)=f(y)^3$. 
So $f(x)=x^n$ or $f(x)=\mathrm{constant}$. 
But I am not getting the given answer if $f(x)=x^n$.

Since $f'(0)$ exists, we know that $f'(0) = \displaystyle\lim_{y \to 0}\dfrac{f(y)-f(0)}{y-0} = \lim_{y \to 0}\dfrac{f(y)}{y}$ exists. 
Then, for any $x \in \mathbb{R}$, we have $\dfrac{f(x+y^3)-f(x)}{y^3} = \dfrac{f(y)^3}{y^3} = \left(\dfrac{f(y)}{y}\right)^3 \to f'(0)^3$ as $y \to 0$. 
Therefore, $f'(x)$ exists for all $x \in \mathbb{R}$ and $f'(x) = f'(0)^3$ for all $x \in \mathbb{R}$. 
Then, since $f(0) = 0$, we have $f(x) = f'(0)^3x$ for all $x \in \mathbb{R}$.
Setting $x = 0$ gives us $f'(0) = f'(0)^3$, and thus, $f'(0) = -1, 0, 1$. 
Since we are given $f'(0) \ge 0$, we must have $f'(0) = 0$ or $1$. 
Thus, $f(x) = 0$ or $f(x) = x$, (both satisfy the constraints of the problem). So there are two possible values for $f(10)$, specifically, $f(10) = 0$ and $f(10) = 10$.
