Q.If $\alpha,\beta,\gamma$ are the roots of $x^3+2x^2-3x-1=0$, find the values of $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$.

Writes the sum of reciprocals as (sum of pairwise products)/(product of roots) and substitutes Vieta's formulas for a cubic.

1. For $x^3+px^2+qx+r=0$ with roots $\alpha,\beta,\gamma$: by Vieta's formulas, $\alpha\beta+\beta\gamma+\gamma\alpha=q$ and $\alpha\beta\gamma=-r$.
2. $\displaystyle\sum\dfrac1\alpha=\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}$.
3. Substitute: $=\dfrac{q}{-r}=-\dfrac qr$.

Using Vieta's formulas for the cubic $x^3+px^2+qx+r$, $\displaystyle\sum\dfrac{1}{\alpha}=-\dfrac{q}{r}$.

1. For $x^3+px^2+qx+r=0$ with roots $\alpha,\beta,\gamma$, Vieta's formulas give: $\alpha+\beta+\gamma=-p$, $\alpha\beta+\beta\gamma+\gamma\alpha=q$, $\alpha\beta\gamma=-r$.
2. We want $\displaystyle\sum\dfrac{1}{\alpha}=\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}$.
3. Combining over a common denominator, $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}=\dfrac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}$.
4. Substituting the Vieta values: numerator $=q$, denominator $=-r$.
5. So $\displaystyle\sum\dfrac{1}{\alpha}=\dfrac{q}{-r}=-\dfrac{q}{r}$.