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n³=(7k+3)³ = (7k)³ + 3*(7k)²*3 + 3*7k*3² + 3³ = 343k³ + 9*49k² + 21k*9 + 27 = 343k³ + 441k² + 189k + 21+6 =7(49k³ + 63k² + 27k + 3) + 6
(49k³ + 63k² + 27k + 3)∈Z, zatem resztą tą jest 6.