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.

Let $f(x) = x^4+x^3-1$ and $g(x) =  x^6+x^4 + x^3 - x^2 - 1$.
Let $a, b, c, d$ be the roots of $f(x)$. It is easy to see these four roots are 
distinct and differ from zero. If we set $\lambda = a + b$ and $\mu = ab$, we will have $\mu \ne 0$. 
Since $a \ne b$ are roots of $f(x)$,  $b$ is a root of 
$$\begin{align}A(x,a) \stackrel{def}{=}
\frac{f(x)-f(a)}{x-a} &= x^3 + (a+1) x^2 + (a^2+a) x + (a^3+a^2)\\
&= x^3 + \frac{x^2}{a^3} + \frac{x}{a^2} + \frac{1}{a}
\end{align}$$
This implies $\mu$ is a root of
$$
B(x,a) \stackrel{def}{=} a^3 A\left(\frac{x}{a},a\right) 
= x^3 + \frac{x^2}{a^2} + x + a^2 = (x^2+1)\left(x+\frac{1}{a^2}\right) - a
$$
Exchange the role of $a,b$, we find $\mu$ is also a root of $B(x,b)$. As a result, $\mu$ is a root of
$$\left(x + \frac{1}{b^2}\right)B(x,a) - \left(x + \frac{1}{a^2}\right)B(x,b)
= -a\left(x + \frac{1}{b^2}\right)
+ b\left(x + \frac{1}{a^2}\right)\\
= \frac{b-a}{a^2b^2}\left[a^2b^2x + (a^2 + ab + b^2)\right]
$$
In terms of $\lambda$ and $\mu$, this leads to
$$\mu^3 + (\lambda^2 - \mu) = 0\quad\iff\quad \lambda^2 = \mu - \mu^3$$
Since $f(a) = f(b) = 0$, we also have
$$\frac{f(a)-f(b)}{a-b} = 
\frac{a^4 - b^4 + a^3-b^3}{a-b} =
(a^2+b^2)(a+b) + (a^2+ab + b^2)  = 0$$
In terms of $\lambda$ and $\mu$, this is equivalent to
$$(\lambda^2 - 2\mu)\lambda + \lambda^2 - \mu = 0
\quad\iff\quad (\mu + \mu^3)\lambda + \mu^3 = 0
$$
Since $\mu \ne 0$, this leads to
$$\mu^4 = (-\mu^2)^2 = (1+\mu^2)^2\lambda^2
= (1+\mu^2)^2(\mu - \mu^3)$$
Get rid of a non-zero $\mu$ from both sides, we get
$$g(\mu) = \mu^6 + \mu^4 + \mu^3 - \mu^2 - 1 = (\mu^2+1)^2(\mu^2 - 1) + \mu^3  = 0$$
This means $ab = \mu$ is a root of the polynomial $g(x)$. 
Swapping the roles of $a,b, c, d$ in suitable order, we can deduce
$ac, ad, bc, bd, cd$ are the other roots of $g(x)$. In short, $g(x)$ has following decomposition:
$$g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd)$$
