The factors of $3{\mathrm{x}}^5-48{\mathrm{x}}^3$ is ____________.

Given, $3{\mathrm{x}}^5-48{\mathrm{x}}^3$

Simplifying by taking common terms, we get
$\Rightarrow 3{\mathrm{x}}^3\left({\mathrm{x}}^2-16\right)$ 

$\Rightarrow 3{\mathrm{x}}^3\left[{\left(\mathrm{x}\right)}^2-{\left(4\right)}^2\right]$ 

Using the identity ${\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\left(a+b\right)\left(a-b\right)$

Here, x = a and y = b

$\Rightarrow 3{\mathrm{x}}^3\left(\mathrm{x}+4\right)\left(\mathrm{x}-4\right)$ 

Hence, factors of $3{\mathrm{x}}^5-48{\mathrm{x}}^3 \mathrm{is}\mathbf{ }\mathbf{3}{\mathbf{x}}^{\mathbf{3}}\left(x+4\right)\left(x-4\right)$. _.