If $\tan\alpha,\, \tan\beta,\,\tan\gamma$ are roots of $au^ 3 +(2a-x)u+ y=0$ If $\tan\alpha,\, \tan\beta,\,\tan\gamma$ are roots of $au^ 3 +(2a-x)u+ y=0$ for fixed $x$ and $y$ and $\tan\alpha + \tan\beta = h$
Find $ah^3 +(2a-x)h$.
Options are
A) $y$
B) $-y$
C) $2a- x$
D) $a$
Solution
We can find
$h= - \tan\gamma$
But I am unable to find correct options.

We have $\tan\alpha+\tan\beta+\tan\gamma=0$
$\implies\tan\gamma=-(\tan\alpha+\tan\beta)=-h$
As $\tan\gamma(=-h)$ is a root of the given equation,  $a(-h)^3+(2a-x)(-h)+y=0$
