$\begin{align} & \text{If }\alpha \text{ and }\beta \text{ are roots of the equation }6{{x}^{2}}-10x+3=0,\text{ } \\ & \text{Find the value of } \\ & \text{(}i\text{) }{{(\alpha -\beta )}^{2}}\text{ } \\ & \text{(}ii\text{) }{{\alpha }^{2}}+{{\beta }^{2}}\text{ } \\ & \text{(}iii\text{) }({{\alpha }^{3}}+{{\beta }^{3}})\text{ } \\ & \text{(}iv\text{) }({{\alpha }^{3}}-{{\beta }^{3}})\text{ } \\ & \text{(}v\text{) }\tfrac{1}{\alpha }+\tfrac{1}{\beta } \\ & (vi)\text{ }\tfrac{1}{{{\alpha }^{2}}}+\tfrac{1}{{{B}^{2}}}\text{ } \\ & (vii)\text{ }\tfrac{\alpha }{\beta }+\tfrac{\beta }{\alpha } \\\end{align}$
$\begin{align}
& \text{Given equation = }6{{x}^{2}}-10x+3=0 \\
& a=6,b=-10,c=3 \\
& \text{Sum of roots}=\alpha +\beta =-\frac{b}{a}=-\frac{10}{6}=-\frac{5}{3} \\
& \text{Product of roots }\alpha +\beta =\frac{c}{a}=\frac{3}{6}=\frac{1}{2} \\
& \text{The value of } \\
& (i)\text{ (}\alpha -\beta {{)}^{2}}={{(\alpha +\beta )}^{2}}-4\alpha \beta ={{\left( -\frac{5}{3} \right)}^{2}}-4\left( \frac{1}{2} \right)=\frac{25}{9}-2=\frac{7}{9} \\
& (ii)\text{ }{{(\alpha +\beta )}^{2}}={{\left( -\frac{5}{3} \right)}^{2}}=\frac{25}{9} \\
& (iii)\text{ }{{\alpha }^{3}}+{{\beta }^{3}}=(\alpha +\beta )\left[ {{(\alpha +\beta )}^{2}}-3\alpha \beta \right] \\
& \text{ }=(-\tfrac{5}{3})\left[ {{(-\tfrac{5}{3})}^{2}}-3(\tfrac{1}{2}) \right]=(-\tfrac{5}{3})(\tfrac{25}{9}-\tfrac{3}{2})=(-\tfrac{5}{3})(\tfrac{23}{18})=-\tfrac{115}{54} \\
& (iv)\text{ }{{\alpha }^{4}}+{{\beta }^{4}}={{\left[ {{(\alpha +\beta )}^{2}}-2\alpha \beta \right]}^{2}}-2{{(\alpha \beta )}^{2}} \\
& \text{ }={{\left[ {{\left( -\frac{5}{3} \right)}^{2}}-2\left( \frac{1}{2} \right) \right]}^{2}}-2{{\left( \frac{1}{2} \right)}^{2}}=\left( \frac{25}{9}-1 \right)-\frac{1}{2}=\frac{16}{9}-\frac{1}{2}=\frac{23}{18} \\
& (v)\text{ }{{\alpha }^{3}}-{{\beta }^{3}}=(\alpha -\beta )\left[ {{(\alpha +\beta )}^{2}}-\alpha \beta \right]=\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }\left[ {{(\alpha +\beta )}^{2}}-\alpha \beta \right] \\
& \text{ }=\sqrt{{{\left( -\frac{5}{3} \right)}^{2}}-4\left( \frac{1}{2} \right)}\left[ {{\left( -\frac{5}{3} \right)}^{2}}-\left( \frac{1}{2} \right) \right] \\
& \text{ }=\sqrt{\frac{25}{9}-2}\left[ \frac{25}{9}-\frac{1}{2} \right]=\sqrt{\frac{7}{9}}\left( \frac{41}{18} \right)=\frac{\sqrt{7}}{3}\times \frac{41}{18}=\frac{4\sqrt{7}}{54} \\
& (vi)\text{ }\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{(\alpha \beta )}^{2}}}=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \beta )}^{2}}} \\
& =\frac{{{\left( -\frac{5}{3} \right)}^{2}}-2\left( \frac{1}{2} \right)}{\left( \frac{1}{2} \right)}=\frac{\tfrac{25}{9}-1}{\tfrac{1}{4}}=\frac{\tfrac{16}{9}}{\tfrac{1}{4}}=\frac{16}{9}\times \frac{4}{1}=\frac{64}{9} \\
& (vii)\text{ }\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta } \\
& =\frac{{{\left( -\frac{5}{3} \right)}^{2}}-2\left( \frac{1}{2} \right)}{\frac{1}{2}}=\frac{\frac{25}{9}-1}{\frac{1}{2}}=\frac{16}{9}\times 2=\frac{32}{9} \\
\end{align}$