How? putting $y = 0$ gives, $\ f(x^4)=x^3f(x) + f(f(0))$ and you didn't prove that $f(f(0)) = 0$.

Mehrab4226 wrote: ↑Wed Dec 23, 2020 7:15 pmOr, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x$ which is indeed the answer.

First of all here $t$ is a positive number and as a consequence, doesn't cover all real values. I mean $f(t) + f(y) \neq f(t + y )$ for all real values of $t$ and $y$.

Finally and most importantly, a

Finally and most importantly, a

**has solution $f(x) = cx$ if the domain is***Cauchy Equation***Rational Number**and if the function's domain is**Real Number**and it fulfils at least one of the following conditions:**(1)**The function is monotonous on some interval**(2)**The function is continuous**(3)**The function is bounded on some interval**(4)**$f(x) \geq 0$ for $x \geq 0$.