Polynomial : $P(a)P(a^2)P(a^3)P(a^4)$ Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.
Find the value of $P(a)P(a^2)P(a^3)P(a^4)$
Is my answer correct ?
Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,
so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$ 
$P(a)P(a^4)=(a^3+2a^2+3a+4)(\frac{1}{a^3}+\frac{2}{a^2}+\frac{3}{a}+4)= 15+5a^4+5a$
Similarly, $P(a^2)P(a^3)=15+5a^3+5a^2$
$P(a)P(a^2)P(a^3)P(a^4)=(15+5a^4+5a)(15+5a^3+5a^2)=125$

Here is another way. Call the product that you want Q.
Note that $xP(x)-P(x)=x^4+x^3+x^2+x-4$
As you correctly observe the values you are substituting are all roots of $x^4+x^3+x^2+x+1=0$ and hence for these values you get $(x-1)P(x)=-5$ whence $$(a-1)(a^2-1)(a^3-1)(a^4-1)Q=(-5)^4=625$$
Now pairing the first and last factor and the middle two factors and noting that $a^5=1$ we have $$(a-1)(a^2-1)(a^3-1)(a^4-1)=(2-a-a^4)(2-a^2-a^3) =$$$$=4-2(a+a^2+a^3+a^4)+a^3+a^4+a+a^2$$ and we know that $$a^4+a^3+a^2+a=-1$$ So $5Q=625$ and $Q=125$
