If $\tan\alpha,\, \tan\beta,\,\tan\gamma$ are roots of $au^ 3 +(2a-x)u+ y=0$

Question

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.

Answer 1

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, $a(-h)^3+(2a-x)(-h)+y=0$