Roots of $ax^2+bx+1 = 0$ are given as $\dfrac{1}{\sqrt\alpha}, \dfrac{1}{\sqrt\beta}$. Find roots of $x^2+(b^3-3ab)x + a^3 = 0$
Method 1:
I assumed $(\alpha)^{-1/2} =x$ and the new variable in which we form the new quadratic as $X$ whose root is $\alpha '$. Then
$$\alpha^{-n/2} = \alpha '\\ \implies x = X^{1/n}$$
Substituting $x$ in first quadratic,
$$aX^{2/n}+bX^{1/n}=-1$$
On cubing
$$a^3X^{6/n}+b^3X^{3/n}+3abX^{3/n}(aX^{2/n}+bX^{1/n}) = -1$$
Substituting $aX^{2/n}+bX^{1/n}=-1$, we get:
$$a^3X^{6/n}+b^3X^{3/n}-3abX^{3/n} = -1$$
Comparing with second quadratic, we get $n=-3$. Thus $$\alpha ' = \alpha^{3/2}$$
Method 2:
I try using sum and product of root formulas:
$$\dfrac{1}{\sqrt\alpha} + \dfrac{1}{\sqrt\beta} = -\dfrac{b}{a}\\ \dfrac{1}{\sqrt\alpha\sqrt\beta} = \dfrac{1}{a}$$
So we get $$\sqrt\alpha + \sqrt \beta = -b$$
So product of roots of neq quadratic is $$\alpha ' \beta ' = a^3 = (\alpha \beta) ^ {3/2}$$
And sum of new roots is $$\begin{align} \alpha ' + \beta ' = 3ab-b^3 &= b(3a-b^2) \\ &= (\sqrt\alpha + \sqrt \beta)((\sqrt \alpha + \sqrt \beta)^2 - 3\sqrt{\alpha\beta} ) \\ &= (\alpha^{3/2} + \beta ^{3/2}) \end{align}$$
Thus comparing gives the same answer.
My question is also that which approach is more proper and general. Also is there an easier approach for this?