<s>[INST] Functional equation: $ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) $ 
Exercise: Find all of functions $f: \mathbb{R} \to \mathbb{R}$ such that $$ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) \text , \forall x , y , z \in \mathbb R \text . $$

In fact, I solved this question as follows:

*

*First step: Calculate $ f ( 0 ) $. In this step, I obtained that $ f ( 0 ) = 0 $ or $ f ( 0 ) = - 1 $.

*Second step: Consider two cases of the first step. I determinated the solutions are $ f ( x ) = 0 $, $ f ( x ) = - 1 $ and $ f ( x ) = x $.

The problem is that this way is very long. Who has another way to solve my exercise?
 [/INST] The case $ f ( 0 ) = 0 $ is solved by @TobEnrack. For the case $ f ( 0 ) = -1 $, let $ x = y = 0 $ in
$$ f \big( x + y ^ 2 + z ^ 3 \big) = f ( x ) + f ( y ) ^ 2 + f ( z ) ^ 3 \tag 0 \label 0 $$
and you get
$$ f \big( z ^ 3 \big) = f ( z ) ^ 3 \tag 1 \label 1 $$
Again, letting $ x = z = 0 $ in \eqref{0} you have
$$ f \big( y ^ 2 \big) = f ( y ) ^ 2 - 2 \tag 2 \label 2 $$
Now combining \eqref{1} and \eqref{2} you get
$$ f ( x ) ^ 6 - 2 = \big( f ( x ) ^ 3 \big) ^ 2 - 2 = f \big( x ^ 3 \big) ^ 2 - 2 = f \Big( \big( x ^ 3 \big) ^ 2 \Big) = f \big( x ^ 6 \big) \\ = f \Big( \big( x ^ 2 \big) ^ 3 \Big) = f \big( x ^ 2 \big) ^ 3 = \big( f ( x ) ^ 2 - 2 \big) ^ 3 = f ( x ) ^ 6 - 6 f ( x ) ^ 4 + 12 f ( x ) ^ 2 - 8 $$
By simple algebra, it's easy to derive
$$ f ( x ) ^ 2 = 1 \tag 3 \label 3 $$
Finally, letting $ z = 0 $ in \eqref{0} and using \eqref{3} you get
$$ f \big( x + y ^ 2 \big) = f ( x ) \tag 4 \label 4 $$
Hence, if $ x \le y $, substituting $ \sqrt { y - x } $ for y in \eqref{4}, it's easy to see that
$$ f ( y ) = f \big( x +  ( y - x ) \big) = f \Big( x + { \sqrt { y - x } } ^ 2 \Big) = f ( x ) $$
Since for every $ x $, either $ x \le 0 $ or $ 0 \le x $, therefore $ f ( x ) = -1 $.
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