At $\left.x=1,\left(x^{3}-1\right)\right]_{x=1}=0$ and $\left.(x-1)\right]_{x=1}=0 $
$\therefore $ values of both branches match at $x=1 $
$ \therefore f(x)$ is continuous at $x=1$
$\left.\left.\left(x^{3}-1\right)^{\prime}\right]_{x=1}=3 x^{2}\right]_{x=1}=3$
and $\left.(x-1)^{\prime}\right]_{x=1}=1 $
$ \therefore $ Slopes of both branches don't match at $x=1 $
$ \therefore f(x)$ is not differentiable