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Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation $f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)\,\forall a,b\in\mathbb R $

Let $$P(x,y): f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)$$ $$P(0,1): f(f(1))=1+f(0)$$ $$P(1,0): f(1+f(0))= f(0)+f(1)$$ Hence, from $P(0,1)$ and $P(1,0)$, we get $$f(1+f(0))=f(0)+f(1) \implies f(0)=f(f(0))=0$$ Now, $$P(0, b): f(f(b^2))=b^2 \quad \forall b \in \mathbb{R^+}$$ What do I do after.

math_learner
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    Many proofs are on an AoPS post here. This is not a duplicate because it is on a different site, but feel free to take and expand on solutions from here, apparently it's an "illustrative" exercise. – Sarvesh Ravichandran Iyer Apr 15 '23 at 11:21
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    You have a typo: it should be $P(a,b)$ rather than $P(x,y)$ in the first line. – MasB Apr 15 '23 at 11:22
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    You have another typo. The mapsto symbol "$\mapsto$" is used to indicate what a function maps a particular input value to, not what its domain and codomain are. For that, an ordinary arrow is used. That is, if $f$ has domain $[1, \infty)$ and codomain $\Bbb R$, and for all $x, f(x) = \sqrt{x-1}$, then you could describe it as $$f : [1,\infty) \to \Bbb R : x \mapsto \sqrt{x-1}$$ or just $f : x \mapsto \sqrt{x-1}$, if the domain and codomain are understood. – Paul Sinclair Apr 16 '23 at 14:04

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