We use, as a starting point, `u = x^3- 2`.

Now `du = 3x^2\ dx`

This time, this is exactly what we have in the question, so there is no need to divide both sides.

We can now write:

`int(x^3-2)^6(3x^2)\ dx`

`=intu^6\ du`

`=u^7/7+K`

`=((x^3-2)^7)/7+K`