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

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

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

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

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

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

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

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