If n is any positive integer, then $(3^{4n}-4^{3n})$ is always divisible by

The given expression is $3^{4n} - 4^{3n}$. This can be rewritten using exponent rules as $(3^4)^n - (4^3)^n$. Calculate the bases: $3^4 = 3 \times 3 \times 3 \times 3 = 81$. Calculate the bases: $4^3 = 4 \times 4 \times 4 = 64$. So, the expression becomes $81^n - 64^n$. A property of exponents states that for any positive integer $n$, $a^n - b^n$ is always divisible by $(a-b)$. In this case, $a = 81$ and $b = 64$. The difference $a-b = 81 - 64 = 17$. Therefore, $81^n - 64^n$ is always divisible by 17.