Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$

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Question:

Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$

My process:

I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.

So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.

Here is the expression I got:

$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$

Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.

And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?

Best Answer

Note that $(x^2+2x+4)(x-2)=x^3-8$. So the product of the $g(a_i)$ is $P/Q$, where

$$P=\prod(a_i^3-8)$$ $$Q=\prod(a_i-2)$$

As @mathcounterexamples.net notes, the roots of $f$ are the cube roots of $4$ and $-2$; so $a_i^3-8$ takes the values $-4,-4,-4,-10,-10,-10$. Hence $P=64000$.

And the product of $a_i-2$ taken over all roots of $f(x)$ is the product of $a_i$ taken over all roots of $$f(x+2)=(x+2)^6-2(x+2)^3-8$$ which is simply the constant term $2^6-2.2^3-8=40$. So we get $P/Q=1600$.

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