# Sum of cubes of three different numbers is equal to the sum of cubes of three other different numbers

I have question when I learned about Ramanujan number I thought are there any number such that cubes of three different numbers is equal to sum of cubes of three different numbers I found an answer for it myself $40^3+30^3+27^3=48^3+4^3+3^3$.

Here my question is it the solution or are there any solutions for it?

• should it be $$a^3+b^3+c^3=d^3+e^3+f^3$$ for some integers $$a,b,c,d,e,f$$? – Dr. Sonnhard Graubner Oct 5 '16 at 14:01
• artofproblemsolving.com/community/c3046h1046691__2 – individ Oct 5 '16 at 14:02
• $29=3^3+1^3+1^3=4^3+(-3)^3+(-2)^3$ – lulu Oct 5 '16 at 14:02
• If you want all positive values, then $251=5^3+5^3+1^3=6^3+3^3+2^3$. – lulu Oct 5 '16 at 14:09
• Lulu I think 5 is repeated so I think you should give another answer but your answer is excellent.how did you find it? – Ch.Siva Ram Kishore Oct 5 '16 at 16:21

\begin{align*} 1009=10^3+2^3+1^3&=9^3+6^3+4^3\\ 1366=11^3+3^3+2^3&=9^3+8^3+5^3\\ 1457=11^3+5^3+1^3&=9^3+8^3+6^3\\ 1520=11^3+5^3+4^3&=10^3+8^3+2^3 \end{align*}
These sums and further examples (though not their decomposition into three distinct positive cubes) are at OEIS. Note, however, that that OEIS sequence includes numbers that are the sum of three distinct positive cubes in different ways where the same cube appears as a term in two ways for the same sum, e.g. $$x^3+1729=x^3+12^3+1^3=x^3+10^3+9^3$$