If α and β are the …

Since,
Sum of the zeroes of polynomial = α + β {tex}= \frac{{ - b}}{a}{/tex}
and product of zeroes of polynomial = αβ {tex}= \frac{c}{a}{/tex}
Simplify the given expression and substitutie the values, we obtain
{tex}\frac{\beta }{{a\alpha + b}} + \frac{\alpha }{{a\beta + b}}{/tex}{tex}= \frac{{\beta \left( {a\beta + b} \right) + \alpha \left( {a\alpha + b} \right)}}{{\left( {a\alpha + b} \right)\left( {a\beta + b} \right)}}{/tex}    
{tex}= \frac{{\alpha {\beta ^2} + b\beta + \alpha^2 {a} + b\alpha }}{{{a^2}\alpha \beta + ab\alpha + ab\beta + {b^2}}}{/tex}
{tex} = \frac{{a{\alpha ^2} + b{\beta ^2} + b\alpha + b\beta }}{{{a^2} \times \frac{c}{a} + ab\left( {\alpha + \beta } \right) + {b^2}}}{/tex}
{tex}=\frac{{a\left[ {{{\left( {\alpha + \beta } \right)}^2} + b\left( {\alpha + \beta } \right)} \right]}}{{ac}}{/tex}
{tex} = \frac{{a\left[ {{{\left( {a + \beta } \right)}^2} - 2\alpha \beta } \right] - \frac{{{b^2}}}{a}}}{{ac}}{/tex}
{tex}= \frac{{a\left[ {\frac{{{b^2}}}{a} - \frac{{2c}}{a}} \right] - \frac{{{b^2}}}{a}}}{{ac}}{/tex}
{tex} = \frac{{a\left[ {\frac{{{b^2} - ac}}{a}} \right] - \frac{{{b^2}}}{a}}}{{ac}}{/tex}
{tex}= \frac{{a\left[ {\frac{{{b^2} - ac - {b^2}}}{a}} \right]}}{{ac}}{/tex}
{tex}= \frac{{{b^2} - 2c - {b^2}}}{{ac}}{/tex}
{tex}= \frac{{ - 2c}}{{ac}} = \frac{{ - 2}}{a}{/tex}