Given equation is, $x^{2}+a x+b=0,(b \neq 0)$
its roots are $\alpha$ and $\beta$. Then, sum of roots $=\alpha+\beta=-a$
Product of roots $=\alpha \cdot \beta=b$
Now,
$$
\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \beta}\right)
$$
$=-a-\frac{(-a)}{b}[$ from Eqs. (0) and $(i i)]$
$=-a+\frac{a}{b}=\frac{a}{b}(1-b)$
and $\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}$
$=b+\frac{1}{b}-2[$ from $E q \cdot(i i)]$
$=\frac{1}{b}\left(b^{2}-2 b+1\right)=\frac{1}{b}(b-1)^{2}$
Required of quadratic equation whose roots are $\left(\alpha-\frac{1}{\beta}\right)$ and $\left(\beta-\frac{1}{\alpha}\right)$ is
$$
\begin{array}{l}
x^{2}-\left\{\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right\} x \\
+\left\{\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)\right\}=0
\end{array}
$$
On putting the values from Eqs. (i) and (ii). we get $x^{2}-\frac{a}{b}(1-b) x+\frac{1}{b}(b-1)^{2}=0$
$\Rightarrow \quad b x^{2}+a(b-1) x+(b-1)^{2}=0, b \neq 0$