If n is a positive integer, prove that

$3^{3 n}-26 n-1=27^{n}-26 n-1 \quad \ldots(1)$

Now, we have :

$27^{n}=(1+26)^{n}$

On expanding, we get

$(1+26)^{n}={ }^{n} C_{0} \times 26^{0}+{ }^{n} C_{1} \times 26^{1}+{ }^{n} C_{2} \times 26^{2}+{ }^{n} C_{3} \times 26^{3}+{ }^{n} C_{4} \times 26^{4}+\ldots{ }^{n} C_{n} \times 26^{n}$

$\Rightarrow 27^{n}=1+26 n+26^{2}\left[{ }^{n} C_{2}+{ }^{n} C_{3} \times 26^{1}+{ }^{n} C_{4} \times 26^{2}+\ldots{ }^{n} C_{n} \times 26^{n-2}\right]$

$\Rightarrow 27^{n}-26 n-1=676 \times$ an integer

Or,

$3^{3 n}-26 n-1$ is divisible by 676 (From (1))