Discussion
If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:
Answer: A
Factorise the expression.
\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]
\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.
Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.
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