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Answer :

Let be a rational number, say where q ≠ 0 and p and q have no common factor.

Taking cube both sides

Since 1 3 = 1 , and 2 3 = 8, it follows that
Then q > 1 because if q = 1 then
will be an integer, and there is no integer between 1 and 2.

being an integer, 6q 2 is an integer, and since q > 1 and q does not have a common factor with p and consequently with p 3 .
is not an integer.

There is contradiction in our assumption therefore
is an irrational number.

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