For the p(x)=ax2+bx+c if α and …

It is given that {tex} \alpha{/tex} and {tex} \beta{/tex} are the zeros of the quadratic polynomial {tex}f(x)=ax^2+bx+c{/tex} 
{tex} \therefore \quad \alpha + \beta = - \frac { b } { a } \text { and } \alpha \beta = \frac { c } { a }{/tex}
Now,

1. {tex}\alpha^3 + \beta ^3{/tex}
   = {tex}(\alpha + \beta )(\alpha^2 -\alpha\beta + \beta ^2){/tex}
   = {tex}(\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta+2\alpha\beta-2\alpha\beta){/tex}
   = {tex} (\alpha + \beta)[(\alpha + \beta)^2-3\alpha \beta]{/tex}
   = {tex}(\alpha + \beta)^3-3\alpha \beta(\alpha + \beta){/tex}
   = {tex} \frac { - b ^ { 3 } +3abc} { a ^ { 3 } }{/tex}
   {tex}=\frac{-b^3+3abc}{a^3}{/tex} 

2. {tex} \frac { 1 } { \alpha ^ { 3 } } + \frac { 1 } { \beta ^ { 3 } } = \frac { \alpha ^ { 3 } + \beta ^ { 3 } } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta) } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta+2\alpha\beta-2\alpha\beta) } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)[(\alpha + \beta)^2-3\alpha \beta] } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { [(\alpha + \beta)^3-3\alpha \beta(\alpha + \beta)] } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex} = \frac { \frac { - b ^ { 3 } +3abc} { a ^ { 3 } } } { \left( \frac { c } { a } \right) ^ { 3 } } {/tex}
   {tex}=\frac{-b^3+3abc}{a^3}\;\times\;\frac{a^3}{c^3}{/tex}
   {tex}= \frac { 3 a b c - b ^ { 3 } } { c ^ { 3 } }{/tex}