- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

- Thread starter
- Admin
- #2

- Feb 14, 2012

- 3,812

$t+u+v=-a,\\tu+uv+vt=b,\\tuv=-c$ and

$t^3+u^3+v^3=-a^3,\\(tu)^3+(uv)^3+(vt)^3=b^3,\\(tuv)^3=-c^3$

Note that

$(t+u+v)^3=t^3+u^3+v^3+3(t+u+v)(tu+uv+vt)-3tuv$

which gives $-a^3=-a^3-3ab+3c$, or equivalently, $c=ab$. In this case $Q(x)$ has the form

$Q(x)=x^3+a^3x^2+b^3x+(ab)^3=(x+a^3)(x^2+b^3)$

This polynomial has a root $x=-a$ and for the other two roots we should have $b\le 0$. Thus the conditions are

$ab=c,\\ b\le 0$