Now this one's interesting. I suppose we can no longer change the value on the right hand side; hence what we could do is place appropriate operations in order to satisfy the equation. It took me awhile to answer all of this but this is what I've come up:

111=6 ---> (1 + 1 +1)! = 6 since this equates to 3!, which is equal to 3x2.

222=6 --> simply add a plus sign in between; hence, 2+2+2 = 6

333=6 --> we can multiply the first two digits then subtract the last one: (3x3)-3 = 6

444=6 --> get the squareroot of all digits (sqrt4 + sqrt4 +sqrt4) = 6

555=6 --> divide the first two digits then add the last one (5/5)+5=6

666=6--> add the first two digits then subtract the last one: (6+6)-6=6

777=6 --> subtract the first digit with the quotient of the last two: 7-(7/7)=6

888=6 --> this one's tricky. get the cuberoot of the product of the first two then add the answer with the cuberoot of the last digit: [cuberoot of(8x8)] + cuberoot of 8

999=6 --> get the square root of the two digits then add this to the last digit raise to the zero power: (sqrt9 + sqrt9 + 9^0) = 6

Math riddles/puzzles are really fun!