A sum of three cubes

Find the positive integer xx with x3=4413+5883+7353x^3 = 441^3 + 588^3 + 735^3.

Show hints (2)+
  1. Each base is a multiple of 147147: 441,588,735=147(3,4,5)441,588,735 = 147\cdot(3,4,5).
  2. Factor out 1473147^3 and use 33+43+53=633^3+4^3+5^3=6^3.

Answer

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882

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Asked at: Jane Street, Optiver

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