A sum of three cubes

EasyBrainteasers~3m

Find the positive integer $x$ with $x^3 = 441^3 + 588^3 + 735^3$ .

Show hints (2)+

Each base is a multiple of $147$ : $441,588,735 = 147\cdot(3,4,5)$ .
Factor out $147^3$ and use $3^3+4^3+5^3=6^3$ .

Answer

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882

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