Part (I) has the counterexample a=b=c(1/2+√5/2). Part (II) does indeed imply a=b=c, which can be proved by taking 2 cases: if a is maximal, a>=b>=c and a>=c>=b

You don't even need cases: the assumption $a=\max\{a,b,c\}$ is enough.

yea wlog let $a$ be greatest than (2) implies $b^4+bc^3=a^4+ab^3>a^4+bc^3$ implies $ b>a$ a contradiction hence $a=b$ and than we easily get $a=b=c$

aditya21 wrote: yea wlog let $a>b>c$ You can't do that. Condition (II) is cyclic, not symmetric.

I did part (II) correct. But in part (I) i got: if $a=b$ then $a=c$ or $c(c^2-a^2)+a^2(a-c)=0$ which i erroneously simplified to $c(c+a)+a^2=0$ and then stated the later was impossible. Therefore a=b=c must hold. How much should I get?

For the sake of completion of part $(1)$ suppose $a=c$ then $a^4 +c^3a=2a^3c$ then$\frac{a^2}{c^2}+\frac{c}{a}-\frac{2a}{c}=0$ consider this as a polynomial in $a/c$ we have $x^3-2x^2+1=0$ $(x-1)(x^2-x-1)=0$ so $c/a$ can assume other values also.

YESMAths wrote: For positive real numbers $a,b,c$ which of the following statements necessarily implies $a=b=c$: (I) $a(b^3+c^3)=b(c^3+a^3)=c(a^3+b^3)$, (II) $a(a^3+b^3)=b(b^3+c^3)=c(c^3+a^3)$ ? Justify your answer. Solution: For $(I)$: we see that, $$ab^3+ac^3=ca^3+cb^3 \implies b^3(a-c)=ca(a^2-c^2) \text{if} a\not =c, then b^3=ca^2+a^2c$$So, for $I$ the condition is not necessary. For $(II)$: Assume that $a=max(a,b,c)$, then we have, $$a^4+ab^3=b^4+bc^3>b^4+ab^3 \implies c^3>ab^2$$$$a^4+ab^3=2b^4>a^4+b^4 \implies b^4>a^4$$Contradiction, so we must have, $a=b=c$.$\blacksquare$

But if $a=\max\{a,b,c\}$ Then $a^3>c^3$ as well as $a>b$, which implies $$a(a^3+b^3) > b(c^3+b^3)$$Then it will be obvious that $a=b=c$?

Note that for (1), $$(a, b, c) = \left( \frac{1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}, 1 \right)$$works. For (II), $b \ge c \iff b \le c$, so $b = c \implies a = b = c$.