Unicode does have superscript versions of the digits 0 to 9: http://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts

This should print 2⁶:

 System.out.println("2⁶");

 System.out.println("2\u2076");

It sounds like, to me, that you are trying to get the 3, representing "cubed", to show as superscript. Displaying text in the superscript or subscript is a formatting task, and as such would require some sort of text manipulation. If your result is going to be displayed in HTML, you can use the tag to cause it to display in the superscript style. As an alternative, you can have the text display as 355 cu-ft, or if you really want the superscript 3, use the unicode representation '\u00B3'. If you open your browser's web console and type '\u00B3', including the quotes, you'll get the superscript 3.

There is no operator, but there is a method.

Math.pow(2, 3) // 8.0

Math.pow(3, 2) // 9.0

---

FYI, a common mistake is to assume 2 ^ 3 is 2 to the 3rd power. It is not. The caret is a valid operator in Java (and similar languages), but it is binary xor.

What do you mean by "display"? If your display device supports Unicode, then use character U+00B3 SUPERSCRIPT THREE , '³'. In Java code you can refer to this character as \u00b3.

Yes. You can do it like this. This is the unicode version for superscript. There are other codes that let you do letters and other digits. This is just an example. Here is a link for the digits and some special characters. I do not know how stable this link is.

System.out.println("km\u00B2");
System.out.println("2\u00B3");

prints

km²
2³

If the target output device supports it, you can also use html.

System.out.println("km<sup>2</sup>")
System.out.println("2<sup>3</sup>")

Update

I have included some code to facilitate multi-digit exponents.

• first, create an array of unicode exponents from 0 to 9 inclusive.

String[] uniDigits = new String[10];        
for(int i = 0; i < 10; i++) {
    uniDigits[i] = switch(i) {
        case 2,3->Character.toString(0xB0+i);
        case 1 -> Character.toString(0xB9);        
        case 0,4,5,6,7,8,9->Character.toString(0x2070+i);
        default -> throw new IllegalArgumentException();       
        };
}

Then I created a BiFunction which takes the exponent and the digit array as arguments and returns the exponent.

BiFunction<Integer, String[], String> exponent = (exp, unis)-> {
    int offset = 0;
    StringBuilder sb = new StringBuilder();
    if (exp < 0) {
        offset = 1;
        exp = -exp;
        sb.append(Character.toString(0x207B)); // minus sign
    }
    int digits = exp == 0 ? 1 : (int)Math.log10(exp)+1;
    while(digits-- > 0) {
        sb.insert(offset, unis[exp%10]);
        exp/=10;
    }
    return sb.toString();
};
    
System.out.println("10%s".formatted(exponent.apply(213,uniDigits)));
System.out.println("15%s".formatted(exponent.apply(-38,uniDigits)));

prints

10²¹³
15⁻³⁸

Instead of using the BiFunction directly, you can also return a Function that takes a single argument of exponent and calls the BiFunction with the exponent and the digit array.

Function<Integer, String> expf = exp-> exponent.apply(exp,uniDigits);

It does the same thing, just saves some typing if you have to sprinkle them around in your code.