Question

If $\alpha ,\beta$ and $\gamma$ are zeros of the polynomial $6x^3+3x^2-5x+1$, then find the value of ${\alpha }^{-1}+{\beta }^{-1}+{\gamma }^{-1}$.

Answer 1

$6x^3+3x^2-5x+1=0$

Here,

$a=6$

$b=3$

$c=-5$

$d=1$

$\because \alpha ,\beta ,\gamma$ are zeroes of given polynomial.

$\therefore \alpha +\beta +\gamma =\frac{-b}{a}=\frac{-3}{6}=\frac{-1}{2}$

$\alpha .\beta .\gamma =\frac{-d}{a}=\frac{-1}{6}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=\frac{-5}{6}$

$\therefore {\alpha }^{-1}+{\beta }^{-1}+{\gamma }^{-1}=\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$

$\Rightarrow =\frac{\beta \gamma +\alpha \gamma +\beta \gamma }{\alpha \beta \gamma }$

$\Rightarrow =\frac{\left(\frac{c}{a}\right)}{\left(\frac{-d}{a}\right)}$

$\Rightarrow =\frac{\left(\frac{-5}{6}\right)}{\left(\frac{-1}{6}\right)}$

$\Rightarrow =5$

$\therefore {\alpha }^{-1}+{\beta }^{-1}+{\gamma }^{-1}=5$