OraQA

The following is an interesting puzzle posted by MathPuzzles :

The number 2008 can be expressed as the sum of 3 positive cubes in exactly 2 different ways. One way is 2^3 + 10^3 + 10^3 = 2008. What is the other way of expressing 2008 as the sum of 3 positive cubes?

———————————SQL Solution—————————-

SELECT DISTINCT rtrim(ltrim(regexp_replace(XMLAgg(XMLElement(X,num)
   ORDER BY num),'<X>|</X><X>|</X>','^3 + '), '^3 + '),' + ') AS OUTPUT
FROM
(SELECT CASE lev
                 WHEN 1 THEN a
                 WHEN 2 THEN b
                 WHEN 3 THEN c END as num,
                rn
FROM
(SELECT a, b, c, rownum rn from
(SELECT LEVEL a from dual CONNECT BY power(LEVEL,3) <=2008) a,
(SELECT LEVEL b from dual CONNECT BY power(LEVEL,3) <=2008) b,
(SELECT LEVEL c from dual CONNECT BY power(LEVEL,3) <=2008) c
WHERE power(a,3) + power(b,3) + power(c,3) = 2008 ),
(SELECT LEVEL lev FROM dual CONNECT BY LEVEL <=3)
)
GROUP BY rn;

OUTPUT
----------------------------------------------
2^3 + 10^3 + 10^3
4^3 + 6^3 + 12^3

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