$\because y=x^3$
Equation of tangent line at point $(\alpha, \beta)$ is:
$(y-\beta)=3 \alpha^2(x-\alpha)$
$\because\left(\alpha_1, \beta_1\right)$
lies on equation (i), we get
$\beta_1-\beta=3 \alpha^2\left(\alpha_1-\alpha\right)$
$\begin{array}{r}\Rightarrow \alpha_1^3-\alpha^3=3 \alpha^2\left(\alpha_1-\alpha\right) \\ \left\{\because\left(\alpha_1, \beta_1\right) \&(\alpha, \beta) \text { lies on } y=x^3\right\}\end{array}$
$\Rightarrow \frac{\alpha_1^3}{\alpha^3}-1=3\left(\frac{\alpha_1}{\alpha}-1\right) \Rightarrow\left(\frac{\alpha_1}{\alpha}\right)^3-3 \frac{\alpha_1}{\alpha}+2=0$
Let $y=\frac{\alpha_1}{\alpha}$
Then, $y^3-3 y+2=0 \Rightarrow(y-1)^2(y+2)=0$
$\begin{aligned} & \Rightarrow\left(\frac{\alpha_1}{\alpha}-1\right)^2\left(\frac{\alpha_1}{\alpha}+2\right)=0 \\ & \because \frac{\alpha_1}{\alpha}-1=0 \text { or } \frac{\alpha_1}{\alpha}=2 \\ & \alpha_1=\alpha \text { (not possible) or } \frac{\alpha_1}{\alpha}=-2 \\ & \Rightarrow\left(\frac{\alpha_1}{\alpha}\right)^3=-8\end{aligned}$
Now, $\frac{\beta_1}{\beta}=\frac{\alpha_1^3}{\alpha^3}=\left(\frac{\alpha_1}{\alpha}\right)^3=-8$