Prove that for all positive integers x and x3 have the same odd/even status. (Hint: proof by cases)

See explanation below

Step-by-step explanation:

Case 1) x is even ⇒ x³ is even

x = 2k for some k

⇒x³ = (2k) ³ = 8k³ = 2 (4k³) which is even.

Case 2) x is odd⇒x³ is odd

x = 2k - 1 for some k

⇒x³ = (2k - 1) ³ = (2k) ³ - 3 (2k) ² (1) + 3 (2k) (1) ² - (1) ³ = 8k³ - 12k² + 6 - 1 = 2 (4k³ - 6k² + 3) - 1 which is odd.