$$\frac{x-t}{x-0}=\frac{y-t^2}{y-0}=\frac{z-t^3}{z-0}=u\text{ (say)}$$
So that $x-t=ux,x=\frac t{1-u}$
Similarly, $y=\frac {t^2}{1-u},z=\frac {t^3}{1-u}$
Clearly, $u\
e1$ as $u=1\implies t=0$
$$\frac{x-t}{x-0}=\frac{y-t^2}{y-0}=\frac{z-t^3}{z-0}=u\text{ (say)}$$
So that $x-t=ux,x=\frac t{1-u}$
Similarly, $y=\frac {t^2}{1-u},z=\frac {t^3}{1-u}$
Clearly, $u\
e1$ as $u=1\implies t=0$