I'm guessing that you mean (x+3)3.
We can write this as (x+3)(x+3)(x+3).
Take the first pair of factors and use the distributive property to expand them:
(x+3)(x+3)=x(x+3)+3(x+3)=x2+3x+3x+9=x2+6x+9.
Now (x+3)(x2+6x+9)=x(x2+6x+9)+3(x2+6x+9)=x3+6x2+9x+3x2+18x+27.
So (x+3)3=x3+6x2+9x+3x2+18x+27=x3+9x2+27x+27.