solution of (3q+1)cube
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0
Answer:
use Euclid's lemma to prove this
Answered by
1
Answer:
x= 3q + 1
x= 3q + 1then, x3 = (3q + 1)3
x= 3q + 1then, x3 = (3q + 1)3= 27q3 + 27q2 + 9q + 1
x= 3q + 1then, x3 = (3q + 1)3= 27q3 + 27q2 + 9q + 1= 9 q (3q2 + 3q + 1) + 1
x= 3q + 1then, x3 = (3q + 1)3= 27q3 + 27q2 + 9q + 1= 9 q (3q2 + 3q + 1) + 1= 9m + 1, where m = q (3q2 + 3q + 1)
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