Question

Let $P\left(n\right)=n^3-n$, the largest number by which $P\left(n\right)$ is divisible $\forall$ possible integral values of $n$ is

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

$P\left(n\right)=n^3-n=(n-1)n(n+1)$
Therefore, $P\left(n\right)$ is the product of $3$ consecutive natural numbers
which is divisible by $3!=6$
Ans: D