Property of 490,689

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Theorem

The number $490 \, 689$ can be expressed as the sum of $3$ cubes in $2$ different ways:

$490 \, 689 = 4^3 + 60^3 + 65^3 = 8^3 + 25^3 \times 78^3$

while at the same time the products of the contributory cube roots of each of those $2$ ways are equal:

$4 \times 60 \times 65 = 8 \times 25 \times 78$


Proof

\(\displaystyle 490 \, 689\) \(=\) \(\displaystyle 64 + 216 \, 000 + 274 \, 625\)
\(\displaystyle \) \(=\) \(\displaystyle 4^3 + 60^3 + 65^3\)
\(\displaystyle 490 \, 689\) \(=\) \(\displaystyle 512 + 15 \, 625 + 474 \, 552\)
\(\displaystyle \) \(=\) \(\displaystyle 8^3 + 25^3 + 78^3\)

Then:

\(\displaystyle 15 \, 600\) \(=\) \(\displaystyle 2^4 \times 3 \times 5^2 \times 13\)
\(\displaystyle \) \(=\) \(\displaystyle 2^2 \times \left({2^2 \times 3 \times 5}\right) \times \left({5 \times 13}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 4 \times 60 \times 65\)
\(\displaystyle \) \(=\) \(\displaystyle 2^3 \times 5^2 \times \left({2 \times 3 \times 13}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 8 \times 25 \times 78\)

$\blacksquare$


Historical Note

This result is reported by David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$ as the work of Stephane Vandemergel, but details are lacking.


Sources