find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ? 
Well I am not able to eliminate or convert $\ x^6$. Please help.

It's not hard to see that $x^4+x^3-1=0$ has two real and two complex roots. If we let these be $a$, $b$, $c+di$, and $c-di$, then we have $a+b+2c=-1$ and $ab(c^2+d^2)=-1$ from the $x^3$ and constant coefficients, $ab+(c^2+d^2)+2c(a+b)=0$ from the (missing) $x^2$ coefficient, and
$${1\over a}+{1\over b}+{1\over c+di}+{1\over c-di}=0$$
from the (missing) $x$ coefficient, which simplifies first to
$${a+b\over ab}+{2c\over c^2+d^2}=0$$
then to
$${a+b\over ab}+(1+a+b)ab=0$$
from which we obtain
$$a+b={-(ab)^2\over1+(ab)^2}$$
Plugging this into the equation $ab+(c^2+d^2)+2c(a+b)=0$, we have
$$ab-{1\over ab}+\left(1-{(ab)^2\over1+(ab)^2} \right){(ab)^2\over1+(ab)^2}=0$$
Writing $p=ab$, this simplifies first to
$${p^2-1\over p}+{1\over1+p^2}\cdot{p^2\over1+p^2}=0$$
Clearing denominators leaves $(p^2+1)(p^4-1)+p^3=0$, which expands out to
$$p^6+p^4+p^3-p^2-1=0$$
Thus $p=ab$ is one solution to $x^6+x^4+x^3-x^2-1=0$
Remark: This answer assumes that where the OP referred to $a$ and $b$ as "the" two solutions of the quartic, they meant the two real solutions, and where they asked for "the" solution of the sextic, they meant to ask for a solution.
