Step 1 :
After factoring out $ 3x $ we have:
$$ 3x^{3}-27x = 3x ( x^{2}-9 ) $$Step 2 :
Rewrite $ x^{2}-9 $ as:
$$ x^{2}-9 = (x)^2 - (3)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 3 $ , we have:
$$ x^{2}-9 = (x)^2 - (3)^2 = ( x-3 ) ( x+3 ) $$