Find the largest prime factor!

$9^6+1=(9^2)^3+1^3$ $=(9^2+1)(9^4+1-9^2)$ $(\small{\color{goldenrod}{Using~a^3+b^3=(a+b)(a^2+b^2-ab)}})$ $=(82)(9^2(9^2-1)+1)=2\times 41 \times 6481$ $\small{\color{#3D99F6}{~~~~~~~~~~~~~~~~~~~~~~~~\space\space\space \text{(Product of three primes)}}}$ Greatest of them= $\Large \color{turquoise}{6481}$

here in this problem we have to find $greatest$ $prime$ $factor$ of the expression: $9^6+1$ $9^6+1=(9^2)^3+1$

now as we know that $a^3+b^3=(a+b)(a^2-ab+b^2)$ so apply this result in above equation, which gives

$81^3+1=(81+1)(81^2-81+1)=82×6481$ which can also be written as $2×41×\boxed{6481}$

so the correct answer is $\boxed{6481}$

Hint: $a^3+b^3=(a+b)(a^2+b^2-ab)$ .

using python :

def max_prime_factor(n):
     largest_factor = 1
     i = 2
     while i * i <= n:
          if n % i == 0:
              largest_factor = i
              while n % i == 0:
                      n //= i
          i = 3 if i == 2 else i + 2
      if n > 1:
           largest_factor = n
      return largest_factor
print max_prime_factor(9**6+1)