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[SOLVED] Cubic equations

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anemone

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Feb 14, 2012
3,812
An equation $x^3+ax^2+bx+c=0$ has three (but not necessarily distinct) real roots $t,\,u,\,v$. For what values of $a,\,b,\,c$ are the numbers $t^3,\,u^3,\,v^3$ roots of an equation $x^3+a^3x^2+b^3x+c^3=0$?
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,812
Let $P(x)=x^3+ax^2+bx+c$ with roots $t,\,u,\,v$ and $Q(x)=x^3+a^3x^2+b^3x+c^3$ whose roots are $t^3,\,u^3,\,v^3$ respectively. By the Viete formula, we have

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

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