How Does Solving P(x) Influence Q(x^2)?

[anemone]

Here is this week's POTW:

-----

Let $P(x)=x^3-2x+1$ and $Q(x)=x^3-4x^2+4x-1$. Show that if $P(r)=0$, then $Q(r^2)=0$.

-----

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. kaliprasad
2. lfdahl

Solution from lfdahl:

Spoiler

Given $P(r) = r^3-2r+1 = 0$:

\[0 = [P(r)]^2 = \left [ r^3-2r+1 \right ]^2 \\\\ = r^6-4r^4+2r^3+4r^2-4r+1 \\\\ = \underbrace{r^6 - 4r^4 + 4r^2-1}_{= Q(r^2)} +\underbrace{2r^3-4r + 2}_{= 2P(r)} = Q(r^2)\;\;\;\; q.e.d.\]