2021/012) If a and b are roots of $x^4 + x^3 - 1$ then prove that ab is a root of $x^6 + x^4 + x^3 - x^2 - 1$

 let the other 2 roots be c and d

then we get comparing coefficients (this is given by viete;s relation also)

a+b+c+d = -1

ab+ac+ad+bc+bd+cd = 0

abc+abd + acd + bdc = 0

abcd = -1

putting a+ b= s c+d = t, ab = p and cd = q we get

s+ t = -1  ...1

p + q + st = 0....2 

pt + qs = 0...3 

and pq = -1...4

$q = - \frac{1}{p}$ (from 4) and t = -1 -s(from 1) we get

$p - \frac{1}{p} - s^2 -s = 0 \cdots(5)$ 

and $p(-1-s) - \frac{s}{p} = 0 => p^(1-s) - s  = 0$

$=> s =\frac{p^2}{1+p^2} \cdots(6)$

from 6 put s in 5 to get

$p - \frac{1}{p} -\frac{p^4}{(p^2+1)^2} +\frac{p^2}{(1+p^2} = 0$

or $p^2(p^2+1)^2 -(p^2+1)^2 -p^5 +p^3(p^2+1) = 0$

or $p^2(p^4+2p^2+1) - (p^4 + 2p^2 + 1)  -p^5 + p^5 + p^3 = 0$

or $p^6 +p^4 +p^3 -p^2 -1 = 0$

so p is a root of $x^6 +x^4 + x^3 -x^2 -1$ 

so ab (which is p)  a root of $x^6 +x^4 + x^3 -x^2 -1$