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.

Easy to see that $ab<0$.
Let $ab=x$.
Thus, we have $$(a^4+a^3)(b^4+b^3)=1$$ or
$$a^3b^3(ab+a+b+1)=1$$ or
$$x^3(x+a+b+1)=1$$ or
$$a+b=\frac{1-x^3-x^4}{x^3}.$$
Also, we have
$$\frac{a^4+a^3-1-(b^4+b^3-1)}{a-b}=0$$ or
$$a^3+a^2b+ab^2+b^3+a^2+ab+b^2=0$$ or
$$(a+b)^3-2ab(a+b)+(a+b)^2-ab=0,$$ which gives $$\left(\frac{1-x^3-x^4}{x^3}\right)^3-2x\cdot\frac{1-x^3-x^4}{x^3}+\left(\frac{1-x^3-x^4}{x^3}\right)^2-x=0$$ or
$$(x^6+x^4+x^3-x^2-1)(x^6-x^4-x^3-x^2+1)=0$$ and since
$x^6-x^4-x^3-x^2+1>0$ for $x<0$, 
we obtain that one of roots it's $ab$ and the second it's $-\frac{1}{ab}.$
The proof that $ab<0$.
Indeed, let $f(x)=x^4+x^3-1$.
Thus, $f'(x)=x^2(4x+3),$ which says that $f$ increases on $\left[-\frac{3}{4},+\infty\right)$ and on $[0,1]$.
But $f(0)f(1)<0$, which says that $f$ has unique positive root.
Also, $f(-1)f(-2)<0$ and since $f$ decreases on $\left(-\infty,-\frac{3}{4}\right],$ 
we see that $f$ has unique negative root.
Id est, $ab<0$.
