3. (BUL 1) Find all polynomials $f(x)$ with real coefficients for which  
$$ f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right) $$
3. An obvious solution is \( f(x) = 0 \). We now look for nonzero solutions. We note that plugging in \( x = 0 \) we get \( f(0)^2 = f(0) \); hence \( f(0) = 0 \) or \( f(0) = 1 \). If \( f(0) = 0 \), then \( f \) is of the form \( f(x) = x^k g(x) \), where \( g(0) \neq 0 \). Plugging this formula into \( f(x) f(2x^2) = f(2x^3 + x) \) we get \( 2^k x^{2k} g(x) g(2x^2) = (2x^2 + 1)^k g(2x^3 + x) \). Plugging in \( x = 0 \) gives us \( g(0) = 0 \), which is a contradiction. Hence \( f(0) = 1 \). For an arbitrary root \( \alpha \) of the polynomial \( f \), \( 2\alpha^3 + \alpha \) must also be a root. Let \( \alpha \) be a root of the largest modulus. If \( |\alpha| > 1 \) then \( |2\alpha^3 + \alpha| > 2|\alpha|^3 - |\alpha| > |\alpha| \), which is impossible. It follows that \( |\alpha| \leq 1 \) and hence all roots of \( f \) have modules less than or equal to 1. But the product of all roots of \( f \) is \( |f(0)| = 1 \), which implies that all the roots have modulus 1. Consequently, for a root \( \alpha \) it holds that \( |\alpha| = |2\alpha^3 - \alpha| = 1 \). This is possible only if \( \alpha = \pm i \). Since the coefficients of \( f \) are real it follows that \( f \) must be of the form \( f(x) = (x^2 + 1)^k \) where \( k \in \mathbb{N}_0 \). These polynomials satisfy the original formula. Hence, the solutions for \( f \) are \( f(x) = 0 \) and \( f(x) = (x^2 + 1)^k, k \in \mathbb{N}_0 \).