Question

Let $\alpha ,\beta ,\gamma$ be three numbers such that $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{1}{2},\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{9}{4}$ and $\alpha +\beta +\gamma =2$, then

Answer 1

(a) ${(\sum \frac{1}{\alpha })}^2=\sum \frac{1}{{\alpha }^2}+2\sum =\frac{1}{\alpha \beta }$
or ${\left(\frac{1}{2}\right)}^2=\frac{9}{4}+2\frac{(\alpha +\beta +\gamma )}{\alpha \beta \gamma }$
or $\frac{1}{4}-\frac{9}{4}=\frac{2.2}{\alpha \beta \gamma }\Rightarrow \alpha \beta \gamma =-2$
$\therefore \left(a\right)\to \left(r\right)$
(b) $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{1}{2}\Rightarrow \frac{\sum \alpha \beta }{\alpha \beta \gamma }=\frac{1}{2}$
$\therefore \frac{\sum \alpha \beta }{-2}=\frac{1}{2}$ by $\left(a\right)\Rightarrow \sum \alpha \beta =-1$
$\therefore \left(b\right)\to \left(s\right)$
(c) $\sum {\alpha }^2=(\sum \alpha {)}^2-2\sum \alpha \beta$
$=(2{)}^2-2(-1)=6$ by $\left(b\right)$
(d) ${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )$
$({\alpha }^2+{\beta }^2+{\gamma }^2-\sum \alpha \beta )$
$\therefore \sum {\alpha }^3-3(-2)=2[-2-(-1\left)\right]=-2$
$\therefore \sum {\alpha }^3=-6-2=-8$.