Show that $41^{n} - 14^{n}$ is a multiple of 27

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Solution

Consider given the given expression $41^{n} - 14^{n}$ ,
$∵ x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + . . . + x y^{n - 2} + y^{n - 1})$
Put, $x = 41$ and $y = 14.$
$∴ 41^{n} - 14^{n} = (41 - 14) (41^{n - 1} + 41^{n - 2} .14 + . . . + 4114^{n - 2} + 14^{n - 1})$
$41^{n} - 14^{n} = 27 (41^{n - 1} + 41^{n - 2} .14 + . . . + 4114^{n - 2} + 14^{n - 1})$

Which is divisible by $27$ ,
Hence, proved

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