find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$

Question

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.

Answer 1

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, $ab<0$.

Answer 2

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:

$$g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd)$$

Answer 3

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.

Answer 4

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.