Show that $2^{3 n} - 1$ is divisible by $7$ for every natural number $n$ .

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Solution

Given
$2^{3 n} - 1$
$= 8^{n} - 1$
$= (8 - 1) [(8^{(n - 1)} + 8^{(n - 2)} + . . . {.8}^{0}]$
we have $8 - 1 - 7$
Therefore 7 is a factor of all the terms in $(8 - 1) [(8^{(n - 1)} + 8^{(n - 2)} + . . . {.8}^{0}]$
Therefore $2^{3 n} - 1$ is divisible by 7 for every natural number of n.

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