Introduction:
To show that a number and its cube leave the same remainder when divided by 6, we need to prove that for any integer 'n', the remainder when 'n' is divided by 6 is the same as the remainder when 'n^3' is divided by 6.

Proof:
Let's consider an integer 'n' and its cube 'n^3'.

Case 1: n is divisible by 6
If 'n' is divisible by 6, then 'n' can be expressed as '6k' for some integer 'k'.
Taking the cube of 'n', we have:
n^3 = (6k)^3 = 6^3 * k^3 = 216k^3

In this case, both 'n' and 'n^3' are divisible by 6, and hence they leave the remainder 0 when divided by 6.

Case 2: n is not divisible by 6
If 'n' is not divisible by 6, then 'n' can be expressed as '6k + r' for some integer 'k' and a remainder 'r' where 0 < r="" />< />
Taking the cube of 'n', we have:
n^3 = (6k + r)^3 = (6k)^3 + 3(6k)^2r + 3(6k)(r^2) + r^3

Expanding the above expression, we have:
n^3 = 216k^3 + 3(36k^2)r + 3(6k)(r^2) + r^3

When we divide 'n^3' by 6, all the terms except 'r^3' are divisible by 6. Hence, the remainder when 'n^3' is divided by 6 is the same as the remainder 'r^3' when 'n' is divided by 6.

Conclusion:
From the above analysis, we can see that for any integer 'n', both 'n' and 'n^3' leave the same remainder when divided by 6. This can be proven using two cases: when 'n' is divisible by 6 and when 'n' is not divisible by 6.