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 $F(x)=x^6+x^4+x^3−x^2−1$. One has 
$$F(x)=(x^2-x+2)f(x)-x^3-x+1\\F(x)=0\iff f(x)=\frac{-x^3-x+1}{x^2-x+2}$$
We look at the values for which $$\frac{-x^3-x+1}{x^2-x+2}=x^4+x^3-1$$
The problem suggests that these values are a simple function of $a$ and $b$. Proving with $a + b$ and with $ab$, this second value is good. In fact
$$\frac{a^3b^3+ab-1}{a^2b^2-ab+2}=a^4b^4+a^3b^3-1$$ it is equivalent to
$$(ab)^6+(ab)^4+(ab^3-(ab)^2-1=0$$
Then $F(ab)=0$  which  (because of $ab(\dfrac{-1}{ab})=-1)$ suggests that $\dfrac{-1}{ab}$ could be another root. It is actually as we can verify.
