What is the equation whose roots are and , if (-^2/) and (-^2/

let our equation based on \(\alpha\) and \(\beta\) be :
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\) -------------------- 1)

We also know that \((-\alpha^{2}/\beta)\) and \((-\beta^{2}/\alpha)\) --------- a)
are roots of the equation \(3x^2–18x+2=0\)

from a) we can conclude that :
Firstly, multiplying them we get, \(\alpha\beta\) = \(\frac{c}{a}\) = \(\frac{2}{3}\).
Secondly adding them we get \(\frac{-\alpha^{2}}{\beta}+\frac{-\beta^{2}}{\alpha} = 6\) => \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} = -6\).
OR
\(\frac{\alpha^{3}+\beta^{3}}{\alpha\beta} = -6\) => substituting the value of \(\alpha\beta\) we get :
\(\alpha^{3}+\beta^{3}= -4\)
here use identity to convert the following into :
\((\alpha+\beta)^{3} - 3*\alpha\beta(\alpha+\beta) = -4\)

substitute value of \(\alpha+\beta=t\) and \(\alpha\beta]\) we get :
\(t^{3} - 2t + 4 = 0.\)

take a glance at the options available to simplify the calculation.

from that we can guess that t is either 2 or -2.
t = -2 satisfies the equation. hence
\( \alpha + \beta = -2\)

Now that we know \(\alpha+\beta\) and \(\alpha\beta\) we can put values in 1) to get
\(3x^{2} + 6x + 2 = 0\)

So IMO A