Proof involving polynomial roots From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.

Considering
\begin{align}
  x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \\
  S_1 &= a+b+c+d \\
  &= -1 \\
  S_2 &= ab+ac+ad+bc+bd+cd \\
  &= 0 \\
  S_3 &= bcd+acd+abd+abc \\
  &= 0 \\
  S_4 &= abcd \\
  &= -1 \end{align}

The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.

Now, using SymmetricReduction[(x-a b)(x-a c)(x-a d)(x-b c)(x-b d)(x-c d),{a,b,c,d}] in Wolfram Alpha or so, we have:
\begin{align}
  f(x) &= (x-\color{red}{ab})(x-\color{red}{ac})(x-\color{red}{ad})
  (x-\color{red}{bc})(x-\color{red}{bd})(x-\color{red}{cd}) \\
  &= x^6-S_2 x^{5}+(S_1 S_3-S_4)x^{4}+(2S_2 S_4-S_3^2-S_1^2 S_4)x^{3} \\
  & \quad +(S_1 S_3 S_4-S_4^2)x^{2}-S_2 S_4^2 x+S_4^3 \\
  &= x^6+x^4+x^3-x^2-1
\end{align}
