Let n be any even number
Let m be any other even number
(Note m could take on the same value as n)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Then if #x# is odd we can write #x -> n+1#
Then #(n+1)^2 -> n^2+2n+1#
Then #(n+1)^3 -> n^3 +3n^2 +3n +1#
There will always be a #1 times 1# which when added to all the even numbers (n's) makes the total odd.
If #x# is divisible by 2 then it is even and #x^6# will also be even.
#color(blue)( x^6 " can be either even or odd")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If #x^2# is odd then let it be represented by #(n+1)#
We have #2x^2# which is #(n+1) + (n+1)# which becomes #2n+2# which is even.
#color(blue)(2x^2 " is always even")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)(3 " is always odd")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So it boils down to the behaviour of combining 3 and #x^6#
Let #x^6 -> (n+1) -> "odd"#
Let #3 -> (m+1) -> "odd"#
#color(red)("Then "x^6 + 3 -> (n+1) + (m+1) = m+n+2" which is even")#
Likewise if n is even #color(red)("then "x^6 + 3 -> (n) + (m+1) = m+n+1" which is odd")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Conclusion:
#f(x)# can be either of odd or even.