Suppose the roots of $ x^4 + x^3 - 1 = 0$ are $ a$, $ b$, $ c$, $ d$. So $ \left\{\begin{array}{lcr} a+b+c+d &= &-1 \\ ab+bc+cd+da+ac+bd &= &0 \\ abc+bcd+cda+dab &= &0 \\ abcd &= &-1 \end{array}\right.$ $ \iff$ $ \left\{\begin{array}{lcr} (a+b)+(c+d) &= &-1 \\ ab+cd+(a+b)(c+d) &= &0 \\ ab(c+d)+cd(a+d) &= &0 \\ (ab)(cd) &= &-1 \end{array}\right.$ And, let $ \left\{\begin{array}{ccl} \alpha &= & a b \\ \beta &= & a+b \\ \gamma &= & c d \\ \delta &= & c+d \\ \end{array}\right.$ So $ \left\{\begin{array}{lcr} \beta+\delta &= &-1 \\ \alpha+\gamma+\beta \delta &= &0 \\ \alpha \delta+\gamma \beta &= &0 \\ \alpha \gamma &= &-1 \end{array}\right.$ Thus $ \left\{\begin{array}{lcr} \delta &= &-1-\beta \\ \gamma &= &-\frac{1}{\alpha} \end{array}\right.$ Hence $ \left\{\begin{array}{lcr} \alpha-\frac{1}{\alpha}+\beta (-1-\beta) &= &0 \\ \alpha (-1-\beta)-\frac{1}{\alpha} \beta &= &0 \\ \end{array}\right.$ So \begin{align*} \alpha (-1-\beta)-\frac{1}{\alpha} \beta = 0 &\iff \alpha (-1-\beta) = \frac{\beta}{\alpha} \\ &\iff \frac{-1-\beta}{\beta} = \frac{1}{\alpha^2} \\ &\iff -\frac{1}{\beta}-1 = \frac{1}{\alpha^2} \\ &\iff \frac{1}{\beta} = -\left(\frac{1}{\alpha^2}+1\right) \\ &\iff \frac{1}{\beta} = -\frac{1+\alpha^2}{\alpha^2} \\ &\iff \beta = -\frac{\alpha^2}{1+\alpha^2} \end{align*} Thus \begin{align*} \alpha-\frac{1}{\alpha}+\beta (-1-\beta)=0 &\iff \alpha-\frac{1}{\alpha}-\frac{\alpha^2}{1+\alpha^2} \left(-1+\frac{\alpha^2}{1+\alpha^2}\right)=0 \\ &\iff \alpha-\frac{1}{\alpha}+\frac{\alpha^2}{(1+\alpha^2)^2}=0 \\ &\iff \frac{\alpha^2(1+\alpha^2)^2-(1+\alpha^2)^2+\alpha^3}{\alpha(1+\alpha^2)^2}=0 \\ &\iff \frac{\alpha^6+\alpha^4+\alpha^3-\alpha^2-1}{\alpha(1+\alpha^2)^2}=0 \end{align*}

Suppose the roots of $ x^4 + x^3 - 1 = 0$ are $ a$, $ b$, $ c$, $ d$. So $\sum_{cyc} a = -1$ $\sum_{cyc} ab = 0 $ $\sum_{cyc} abc = 0 \iff \sum_{cyc}\frac{1}{a} = 0$ $abcd = -1$ Suppose the roots of $ x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$ are $ab$, $bc$, $cd$, $da$, $ac$, $bd$. \begin{align*} -a_5 &= \sum_{cyc} ab \\ &= 0 \end{align*} \begin{align*} a_4 &= abcd\left(\sum_{cyc} \frac{a+b+c}{d} + 3\right) \\ &= abcd\left(\sum_{cyc} \frac{1}{a} - 1\right) \\ &= (-1) \times (0-1) \\ &= 1 \end{align*} \begin{align*} -a_3 &= abcd\left(\sum_{cyc} a^2 + 2\sum_{cyc} ab + \sum_{cyc}\frac{abc}{d}\right) \\ &= abcd\left(\left(\sum_{cyc}a\right)^2-22\sum_{cyc} ab\right) \\ &= (-1) \times ((-1)^2-2 \times 0) \\ &= -1 \end{align*} \begin{align*} a_2 &= (abcd)^2\left(\sum_{cyc} \frac{a+b+c}{d} + 3\right) \\ &= (abcd)^2\left(\sum_{cyc} \frac{1}{a} - 1\right) \\ &= (-1)^2 \times (0-1) \\ &= -1 \end{align*} \begin{align*} -a_1 &= (abcd)^2\sum_{cyc} ab \\ &= (-1)^2 \times 0 \\ &= 0 \end{align*} \begin{align*} a_0 &= (abcd)^3 \\ &= (-1)^3 \\ &= -1 \end{align*} Therefore $(a_5, a_4, a_3, a_2, a_1) = (0, 1, 1, -1, 0, -1)$. Thus $x^6 + x^4 + x^3 - x^2 - 1 = 0$.

Let $k = a+b$, $m = ab$. Let the other two roots of $x^4 + x^3 - 1 = 0$ be $c, d$. Then \begin{align*} x^4 + x^3 - 1 &= (x-a)(x-b)(x-c)(x-d) \\ &= x^4 - (a+b+c+d)x^3 + (ab + bc + cd + da + ac + bd)x^2 - (abc + bcd + cda + dac)x + abcd, \end{align*} so by comparing coefficients, we have $a+b+c+d = -1$ and $abcd = 1$. Consequently, $c+d = -1-k$, and $cd = 1/m$, and it follows that \begin{align*} x^4 + x^3 - 1 &= (x^2 - (a+b)x + ab)(x^2 - (c+d)x + cd) \\ &= (x^2 - kx + m)(x^2 +(k+1)x -1/m) \\ &= x^4 + x^3 + (m - m^{-1} - k(k+1))x^2 + ((k+1)m - km^{-1})x - 1. \end{align*} This then gives the additional relationships \begin{align*} 0 &= m^2 - 1 - k(k+1)m, \\ 0 &= k(m-m^{-1})+m. \end{align*} Solving the latter equation gives $k = \frac{-m^2}{m^2+1} = -1 + \frac{1}{m^2+1}$, where upon substituting into the former gives $0 = m^2 - 1 + \frac{m^3}{(m^2+1)^2}$. Multiplying through by the denominator then gives \begin{align*} 0 &= (m^2+1)^2 (m^2 - 1) + m^3 \\ &= m^6 + m^4 + m^3 - m^2 - 1, \end{align*} which proves that $m = ab$ is a root of the polynomial $x^6 + x^4 + x^3 - x^2 - 1$ $\Box$.

3000th post Let $a,b,c,d$ be the roots of the original equation. Then $ab, ac, ad, bc, bd, cd$ are the roots of the next equation is what we're trying to prove. Note that: $S_0=1=a_6$. $S_1=ab+ac+ad+bc+bd+cd=0=-a_5$ $S_2=3abcd+\sum(a^2bc+ab^2c+abc^2)=3abcd+\sum abc(a+b+c)=3abcd+\sum abc(-1-d)=-abcd-abc-abd-acd-bcd$. Note that $abcd=-1$, and $abc+abd+acd+bcd=0$ (both from previous equation), so we get $1=a_4$ $S_3=abcd(a^2+b^2+c^2+d^2)+\sum a^2b^2c^2+\sum 2a^2b^2cd=-1*1+ (abc+abd+acd+bcd)^2-2(\sum a^2b^2cd)+2\sum 2a^2b^2cd=-1$ so we get the coefficient is $(-1)^3(-1)=1=a_3$. $S_4=abcdS_2=-1$ so the coefficient of this is $-1=a_2$ $S_5=a^2b^2c^2d^2S_1=0$ so the coefficient of this is $0=a_1$. $S_6=a^3b^3c^3=-1$ so the coefficient of this is $-1=a_0$. Therefore, the polynomial with roots $ab, ac, ad, bc, bd, cd$ is $a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=x^6+x^4+x^3-x^2-1$ so we are done $\Box$

A USO 1977 problem

Denote the four roots by $a$, $b$, $c$, $d$. Then, we have by Vieta \begin{align*} -1 &= (a+b) + (c+d) \\ 0 &= (a+b)(c+d) + ab + cd \\ 0 &= ab(c+d) + cd(a+b) \\ -1 &= ab \cdot cd. \end{align*}Let $s = a+b$, $p = ab$. Then $c+d = -1-s$ and $cd = -1/p$ from the first and last equation. Substituting these into the center two equations gives: \begin{align*} 0 &= s(-1-s) + p - \frac 1p \\ 0 &= p(-1-s) - \frac sp. \end{align*}The latter equation allows us to solve for $s$ in terms of $p$, and we get $s = \frac{-p}{p+1/p} = \frac{-p^2}{p^2+1}$. Finally, plugging into the former, \begin{align*} 0 &= \frac{-p^2}{p^2+1} \cdot \left( -1 + \frac{p^2}{p^2+1} \right) + p - \frac 1p = \frac{p^2}{(p^2+1)^2} + \frac{p^2-1}{p} \\ &= \frac{p^3 + (p^2-1)(p^2+1)^2}{p(p^2+1)^2} = \frac{p^6+p^4+p^3-p^2-1}{p(p^2+1)^2} \end{align*}as desired.

$x^4+x^3-1=(x-a)(x-b)(x^2+cx-\frac{1}{ab})$ for certain $c\in R$. Comparing coefficients on both sides we get $-a-b+c=1\wedge ab-(a+b)c-\frac{1}{ab}=0\wedge abc+\frac{a+b}{ab}=0\implies 1+a+b=\frac{ab-\frac{1}{ab}}{a+b}=\frac{-a-b}{(ab)^2}=c$ Hence $a+b=\frac{-(ab)^2}{1+(ab)^2}$ Plug this into $1+a+b=\frac{ab-\frac{1}{ab}}{a+b}$ and get the answer.

Here is my solution for this problem Solution Let $c$, $d$ be the rest $2$ roots of $x^4 + x^3 - 1 = 0$ then $abcd = - 1$ and $a + b + c + d = - 1$ We need to prove that: $a^6b^6 + a^4b^4 + a^3b^3 - a^2b^2 - 1 = 0$ or $a^3b^3 + ab + 1 + cd + c^3d^3 = 0$ We have: $a^4 + a^3 - 1 = 0$ so: $a^3 = \dfrac{1}{a + 1}$ Similarly: $b^3 = \dfrac{1}{b + 1}$ Then: $a^3b^3 = \dfrac{1}{(a + 1)(b + 1)} = - (c + 1)(d + 1)$ Similarly: $c^3d^3 = - (a + 1)(b + 1)$ Hence: $a^3b^3 + ab + 1 + cd + c^3d^3 = - (c + 1)(d + 1) + ab + 1 + cd - (a + 1)(b + 1) = - 1 - a - b - c - d = 0$ or $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1 = 0$

Because P(x) has only two real roots, so all the solutions, except one of WolfusA, is using complex numbers.

Assume the roots of the equation $x^4+x^3-1=0$ are $\alpha ,\beta ,\gamma ,\delta$ We have to prove that $\alpha\beta$ are the roots of the equation $x^6+x^4+x^3-x^2-1=0.$ By vietas relation we get $$(\alpha+\beta)+(\gamma+\delta)=-1$$$$\alpha\beta+\gamma\delta+(\alpha+\beta)(\gamma+\delta)=0$$$$\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)=0$$$$(\alpha\beta)(\gamma\delta)=-1$$denote $(\alpha+\beta)=s$ $(\gamma+\delta)=s`$ $\alpha\beta=p$ $\gamma\delta=p`$ putting this value on given equations we get $$(\alpha+\beta)+(\gamma+\delta)\implies\boxed{s+s`=0}$$$$\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)\implies\boxed{ps`+p`s=0}$$$$(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta\implies\boxed{p+p`+ss`=0}$$$$(\alpha\beta)(\gamma\delta)\implies\boxed{pp`=0}$$From sum of roots we get $\boxed{s`=-1-s}$ and from product of roots we get $\boxed{p`=\frac{-1}{p}}$ putting this value on given equations we get $$\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)\implies\boxed{s=-\frac{-p^2}{p^2+1}}$$from these we also get $$\boxed{s`=\frac{-1}{p^2+1}}$$Putting all values in sum of roots taken two at a time we get $$(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta\implies p-\frac{1}{p}+\frac{p^2}{(p^2+1)^2}=0\implies\boxed{p^6+p^4+p^3-p^2-1=0.}$$Proved... Vietas Theorem!!!!