Question

please help me in this question​

Answer 1

first find the zeros of given polynomial and than put the values of

$1 \div \alpha + 1 \div \beta$

Answer 2

$5 {y}^{3} - 7y + 1 \\ \alpha + \beta = \frac{7}{5} \\ \alpha \beta = \frac{1}{5} \\ \frac{1}{ \alpha } \ + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } \\ \frac{7}{5} \div \frac{1}{5} = 7$