<s>[INST] let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$ Question: let P(x) is a polynomial function
Given $P(x^2)= x^2(x^2+1) P(x)$ and 
$P(2)=3$, Find $P(3)$
If I put $x=2$, we get $P(4)$ which is not required.
I tried to find $P(x)$ to let it general polynomial to find its roots but unable to find
Please help!
 [/INST] Elaborating on some details:
$P(0)=P(0^2)=0^2(0^2+1)P(0)=0$
$P(1^1)=P(1)=1^2(1^2+1)P(1)$ so $P(1)=2P(1)$ implying $P(1)=0$
$0=P(1)=P((-1)^2)=(-1)^2((-1)^2+1)P(-1)=2P(-1)$ implying $P(-1)=0$
So, we have found three roots so far.

As for the degree of $P$, if we were to suppose $P(x)$ were a degree $n$ polynomial, it follows that $P(x^2)$ should be a degree $2n$ polynomial.
As $\underbrace{P(x^2)}_{\text{degree }2n}=\underbrace{x^2(x^2+1)}_{\text{degree }4}\times \underbrace{P(x)}_{\text{degree }n}$ it follows that $2n=4+ n$ and so $n=4$

By the fundamental theorem of algebra, it follows that $P(x)$ can be written in the form:
$$P(x)=x(x-1)(x+1)(ax+b)$$
for some values $a,b$
Let us try to find out those values of $a,b$ using $P(2)$ and $P(4)$.
From $P(2)$ we get that $3=P(2)=2(2-1)(2+1)(2a+b)$ and so $2a+b=\frac{1}{2}$.  That is a good start but not quite enough to finish, so we use one more.
From $P(4)$ we get $P(4)=P(2^2)=2^2(2^2+1)P(2)=4\cdot 5\cdot 3 = 60$ and that $P(4)=4(4-1)(4+1)(4a+b)$ and so $4a+b=1$
Subtracting the first result away from the second, this gives $2a=\frac{1}{2}$ and $a=\frac{1}{4}$, thus $b=0$
As a result, we have that $$P(x)=x(x-1)(x+1)(\frac{1}{4}x+0)$$ or rewritten as you had arrived at in the comments $$P(x)=\frac{x^2(x^2-1)}{4}$$
From here, finding $P(3)$ is a matter of just plugging in the value and we get $P(3)=\frac{9\cdot 8}{4}=18$
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