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

Answer 1

Cube the integer :
(a + 1/a)^3 = a^3 + 3a + 3/a + 1/a^3

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

Now rearrange to find that :
a^3 + 1/a^3 = (a + 1/a)^3 - 3(a + 1/a)

= integer^3 - 3 * integer

= an integer

Answer 2

For a+1/a to be an integer, a must be equal to 1.

If the value of 'a' is an integer other than 1, then 1/a will be a fraction.
Therefore, a+1/a will be a fraction.

If the value of 'a' is a fraction, then 1/a will be an integer.
Therefore, a+1/a will be a fraction.

So, a has to be 1.

With a=1,
a^3+1/a^3
=1^3+1/1^3
=1+1/1 [because 1^(any number) is always equal to 1]
=1+1
=2, which is an integer.

Answer 3

if you cube a+1/a you get
a^3 + 3*a +3/a +1/a^3
as a+1/a is integer its cube is also integer
therefore (a+1/a)^3 -3(a+1/a) is also integer

that is same as a^3 +1/a^3.
hence a^3+ 1/a^3 is also integer.

Answer 4

wel in the expression
a + 1/a

if a>1 1/a wil be a fraction
if a<1 a is a fraction

so the expression
a + 1/a will be an integer only for a=1 or -1

hence accordingly even
a^3 + 1/a^3 will be an integer for a=1 or -1

hence the condition is same for both exp to be equal

Answer 5

integers = any whole number including negatives


if a = 1

1+ 1/1 = 2<-integer
1^3 + 1 / 1^3 = 2 <-integer


a can also = -1

Answer 6

(a+1/a)^3= a^3+ 1/a^3 + 3a*1/a^2+3*a^2*1/a=
(a+1/a)^3= a^3 + 1/a^3 +3*(a+1/a) so
integer= a^3+1/a^3+ integer so
a^3 +1/a^3 =integer