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 Cauchy Equation has solution $f(x) = cx$ if the domain is 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 $x \geq 0$.
Finally and most importantly, a Cauchy Equation has solution $f(x) = cx$ if the domain is 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 $x \geq 0$.