Answer :

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

Then

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 } .

So, is not an integer.

Thus

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

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