2021/085) Let $P(x)=x^3−2x+1$ and $Q(x)=x^3−4x^2+4x−1$. Show that if P(r)=0, then $Q(r^2)=0$

 We have

$P(r) = r^3 - 2r + 1= 0\cdots(1)$

and $Q(r^2) = r^6 - 4r^4 + 4r^2 -1 \cdots(2)$

as $r^3 = 2r  - 1$ so $r^6 = (2r-1)^2 = 4r^2 - 4r + 1$
Putting in (1)  

$Q(r^2) = 4r^2 -4r + 1 - 4r^4 + 4r^2 - 1 = -4r^4 + 8r^2 - 4r = -4r(r^3 - 2r + 1) = 0$ using (1)