Prove that if $a$ is an integer, then $a^{3} - a$ is divisible by $6$ .Also deduce that $5 a^{3} + 13 a - 30$ is divisible by 6.

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Solution

$\begin{matrix} a^{3} - a = & a (a^{2} - 1) = a (a - 1) (a + 1) = (a - 1) \cdot a \cdot (a + 1) \end{matrix}$

3 consecutive integers are always divisible by 6
thus 6 divides $a^{3} - a \Rightarrow a^{3} - a = 6 m$

$5 a^{3} + 13 a - 30$
$\begin{matrix} = 5 (a^{3} - a) + 18 a - 30 = 5 (a^{3} - a) - 6 (3 a - 5) = 5.6 m - 6 \cdot (3 a - 5) = 6 (5 m - 3 a - 5) \end{matrix}$

$\Rightarrow 6$ divides $5 a^{3} + 130 - 30$

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Similar questions

1. Prove that the product of two consecutive positive integers is divisible by 2.

2. Prove that the product of three consecutive integers is divisible by 6.

FOR ANY POSITIVE INTEGER n , prove that n³- n is divisible by 6.

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