Factorize (k^3 - k) to get k(k^2 - 1) = k (k + 1) (k - 1)

Therefore the 4 terms are (k - 1) * k * (k + 1) (k + 2) which are 4 consecutive numbers...and the product of any 4 consecutive numbers is always divisible by 6.

Therefore the remainder is always 0

IMO A

Given that k is a positive integer and we need to find what is the remainder when \((k + 2)(k^3 – k)\) is divided by 6

Let's simplify \((k + 2)(k^3 – k)\)

\((k + 2)(k^3 – k)\) = \((k + 2)(k*(k^2 – 1))\) = k*(k+2)*\((k^2 - 1^2)\) = k*(k+2)*(k-1)*(k+1) = (k-1)*k*(k+1)*(k+2)
= Product of four consecutive numbers

Theory: Product of n consecutive numbers will always be divisible by n!

=> Product of four consecutive numbers will always be divisible by 4! (=1*2*3*4 = 24)
=> Product of four consecutive numbers will always be divisible by 6, as 6 is a factor of 24

=> The remainder when \((k + 2)(k^3 – k)\) is divided by 6 = 0

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Remainders