Factorise the given expression          (1)3n(2n-3+m)2n(3n-9+3m)3(2n-3n+m)

The given expression is: 3n(2n-3+m)2n(3n-9+3m)3(2n-3n+m)

First, let's simplify the expression inside each bracket:

3n(2n-3+m) becomes 3n(2n+m-3) 2n(3n-9+3m) becomes 2n(3n+3m-9) 3(2n-3n+m) becomes 3(n+m-3n)

Now, let's distribute the terms outside the brackets:

3n(2n+m-3) becomes 6n^2+3mn-9n 2n(3n+3m-9) becomes 6n^2+6mn-18n 3(n+m-3n) becomes 3n+3m-9n

So, the expression becomes: 6n^2+3mn-9n+6n^2+6mn-18n+3n+3m-9n

Now, let's combine like terms:

6n^2+6n^2 becomes 12n^2 3mn+6mn becomes 9mn -9n-18n+3n-9n becomes -33n 3m is the only term of its kind

So, the final expression is: 12n^2+9mn-33n+3m

This expression cannot be factorised further as there are no common factors among the terms.