If α, β, γ are the zeroes of the polynomial
Question:

If $\alpha, \beta, y$ are the zeroes of the polynomial $p(x)=6 x^{3}+3 x^{2}-5 x+1$, find the value of $\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)$

Solution:

Given : $p(x)=6 x^{3}+3 x^{2}-5 x+1$

$=6 x^{2}-(-3) x^{2}+(-5) x-(-1)$

Comparing the polynomial with $x^{3}-x^{2}(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma$, we get:

$\alpha \beta+\beta \gamma+\gamma \alpha=-5$

and $\alpha \beta \gamma=-1$

$\therefore\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)$

$=\left(\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}\right)$

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

$=5$