If 1 and 1 are the roots of ax^2+bx+1=0 (a0, a,b R), then x(x+b^3)+(a^3-3abx)=0 has roots:

Question

If $\dfrac1{\sqrt\alpha}$ and $\dfrac1{\sqrt\beta}$ are the roots of $ax^2+bx+1=0$ ($a
e0$, $a,b\in\mathbb R$), then $x(x+b^3)+(a^3-3abx)=0$ has roots:

Accepted Answer

$\dfrac1{\sqrt\alpha}+\dfrac1{\sqrt\beta}=-\dfrac ba$, $\dfrac1{\sqrt{\alpha\beta}}=\dfrac1a \implies a=\sqrt{\alpha\beta}$, $b=-(\sqrt\alpha+\sqrt\beta)$.

Substituting into $x^2+(b^3-3ab)x+a^3=0$ and simplifying (using symmetric function identities) shows the roots are exactly $\alpha^{3/2}$ and $\beta^{3/2}$.

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