Let $p(x)=x^2-x+1$. Let $\alpha$ be a root of $p(p(p(p(x))))$. Compute $|(p(\alpha)-1)p(\alpha)p(p(\alpha))p(p(p(\alpha)))|$.

Question

I feel like I'm supposed to use complex numbers for this question, though I'm not exactly sure how to do so.

I suspect that I'm supposed to use the factorization $(x^3+1)=(x+1)(x^2-x+1)$. However, because I'm finding a root of $p(p(p(p(x))))$, I'm not quite sure how to use this without bashing.

The only possibly relevant progress I've made is to notice that $(p(p(p(\alpha))))^2-(p(p(p(\alpha))))+1=0$, yielding $(p(p(p(\alpha))))=e^{\pm \frac{i\pi}{3}}$.

Answer 1

It can be thought in following way,

$$h(x)=(p(x)-1)*p(x)*(p(p(x))*p(p(p(x)))$$ $$h(x)=([p(x)]^2-p(x))*p(p(x))*p(p(p(x)))$$

If you notice $[p(x)]^2-p(x)$ add and subtract 1

So, $[p(x)]^2-p(x)+1-1=p(p(x))-1$

SO, $$h(x)=[p(p(x)-1]*p(p(x))*p(p(p(x)))$$ $$h(x)=[[p(p(x)]^2-p(p(x))]*p(p(p(x)))$$

Now similarly do same again, $[p(p(x))]^2-p(p(x))+1-1=p(p(p(x)))-1$

$$h(x)=[p(p(p(x)))-1]*p(p(p(x)))$$

And one more time ,SO

$$h(x)=[p(p(p(p(x))))]-1$$

Now $\alpha$ is root of $p(p(p(p(x))))$ so it will be zero

$$|h(\alpha)|=|-1|=1$$