If: a + 1/a is an integer, prove that a^3 + 1/a^3 is an integer.?

Question

Please well explained.

Answer 1

Nice. Follow these steps. a + 1/a is an integer, therefore

(a + 1/a)*3 is an integer

a^3 + 3a + 3(1/a) + 1/a^3 is an integer

But 3(a + 1/a) is an integer

Therefore, subtracting the two, we see that

a^3 + 1/a^3 is an integer.

Answer 2

Since a + 1/a is an integer, then (a + 1/a) ^3 is an integer.

Also 3 * (a + 1/a) is an int. And because the difference of two ints is an int, (a+1/a)^3 - 3*(a+1/a) is an int

(a + 1/a)^3 = a^3 + 3a + 3/a + 1/a^3
3 *(a+1/a) = 3a +3/a

So (a+1/a)^3 - 3*(a+1/a) = a^3 + 1/a^3 is an integer.

Answer 3

If (a +1/a) is an integer then so is any integral power and so
(a + 1/a) ^3 is an integer,

Expanding you have
(a + 1/a) ^3 = a^3 + 3a^2x1/a + 3a x 1/a^2 + (1/a)^3 = a^3 +(1/a)^3 + (3a + 3/a) = a^3 + (1/a)^3 +3(a +1/a) and since (a +1/a) is given as an integere so is 3(a +1/a) and so is a^3 + (1/a)^3 = (a + 1/a) ^3 - 3(a +1/a) , the difference of 2 integers being also an integer!