Let n be any even number

Let m be any other even number

(Note m could take on the same value as n)

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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")#

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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")#

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#color(blue)(3 " is always odd")#

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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")#

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Conclusion:

#f(x)# can be either of odd or even.