Polynomial such that $P(x^3) = P(1-x^3)$. Find all monic polynomials of degree $10$ such that $P(x^3) = P(1-x^3)$ for all real $x$ and the constant coefficient is $1$.

The constant coefficient is $1$, since $P(0)=1.$ I have tred setting up a system of equations but it seems to tedious. Is there a slick way to do this?
 A: $t \in \mathbb{R} \Rightarrow P(t) = P(1 - t)$.
Hence if $\alpha$ is a root of $P(x)$ so is $1 - \alpha$.
Thus pick any 5 complex numbers $a_1 , a_2, a_3, a_4, a_5$ satisfying

*

*$\prod_{i = 1}^5 a_i (1 - a_i) = 1$

*$\forall i \; \exists $j$ \; a_i = \bar{a_j} \lor a_i = 1 - \bar{a_j}$
Then:
$$P(x) = \prod_{i = 1}^5 (x - a_i)(x - 1 + a_i)$$
P.S: It doesn't matter if any $a_i$ is $\frac{1}{2}$, a double root must ensure in this case.
Condition 2 evaluates to: $a_i$ is either
(1) Real
(2) Non-real, but has real part $\frac{1}{2}$
(3) Non-real, does not have real part $\frac{1}{2}$ but has a complex conjugate pair.
(4) Non-real, does not have real part $\frac{1}{2}$ , does not have a complex conjugate pair, but there exists another $a_j$ which is the reflection of $a_i$ at the point $(\frac{1}{2}, 0)$ in the Argand plane.
A: Suppose you have a $P$ that satisfies the conditions.
As already noted, $P(t)=P(1-t)$, so if we define
$$ Q(x) = P\bigl(x+\tfrac12\bigr) $$
then $Q$ is an even funtion. Also $Q$ is still a monic polynomial of degree $10$, but being even means that the odd-degree coefficients must all vanish. We also know $Q(-\tfrac12)=1$. All in all $Q$ must have the form
$$ Q(x) = x^{10}+ax^8+bx^6+cx^4+dx^2+\Bigl(1-\frac{1}{2^{10}}-\frac{a}{2^8}-\frac{b}{2^6}-\frac{c}{2^4}-\frac{d}{2^2}\Bigr) $$
Conversely, every $Q$ of this shape gives rise to a valid $P$ when we define
$$ P(t) = Q\bigl(t-\tfrac12\bigr) $$
Now all there's left is a tedious but (if you remember the binomial theorem) straightforward maze of basic algebra if you want precise expressions for the coeffients of $P$ as functions of $a,b,c,d$.
