Question: ```
  firstLetter = word.charAt(0);
  lastLetter = word.charAt((word.length()) - 1);
  noFirstLetter = word.substring(1);
  newWord = noFirstLetter + firstLetter;
  if(word.equalsIgnoreCase(newWord))

```

So im trying to take the first letter of the word and if I take the first letter of the word away and move it to the end it should be equal to the same word. My code here isn't working. For example if the user entered "dresser" if you move the "d" to the end of the word you get the word "dresser" again. That is what im trying to check

Answer: 
I think what you are trying to do is to remove the first character, and then check if rest of the characters are symmetrical around the center of the word.(OK even it is complicated for me)

EG:

dresser => (drop d) => resser => (add d to the right) => resserd (read from right to left and it is dresser again). 

after you drop the first letter:

resser (there are even number of letters in the word)

```
r e s s e r
    |_|
  |_____|
|_________|

```

since they are symmetrical, you can say that that word would be Palindromic if you move D from left to right (or vice versa).

The whole thing above could be horribly wrong if I misunderstood your question at the first place. But I assumed, your question was NOT a simple substring question.

Cheers.
If you move 'd' to the end of the word "dresser" you get "resserd" which is not equal to "dresser".

If you move 'd' to the end and then **reverse** the word then they are equal.

So, assuming you consider strings equals even if they are reversed, the code you want would be :

```
boolean testPass = false;
if ( word.length()==0 )
  testPass = true;
else
{
  firstLetter = word.charAt(0);
  noFirstLetter = word.substring(1);
  newWord = noFirstLetter + firstLetter;
  reversed = new StringBuilder(newWord).reverse().toString()
  if (word.equalsIgnoreCase(newWord) || word.equalsIgnoreCase(reversed))
    testPass = true;
}
if ( testPass )
  // Do something

```

Take notice of the important check of word having lenght 0. Otherwise word.charAt(0) will throw an exception.