Factorise:

We have:

$3 x^{5}-48 x^{3}=3 x^{3}\left(x^{2}-16\right)$

$=3 x^{3}\left\{(x)^{2}-(4)^{2}\right\}$

$=3 x^{3}(x+4)(x-4)$

$\therefore 3 x^{5}-48 x^{3}=3 x^{3}(x+4)(x-4)$