Hi Blisstry Mystery,

$$a^m-b^m$$ is always divisible by (a-b) where m is a positive integer

For example - $$a^2-b^2$$ is divisible by a-b

$$a^3-b^3$$ is divisible by a-b

$$a^4-b^4$$ is divisible by a-b

Therefore, here $$30^{65}-29^{65}$$ is in the numerator where a = 30 and b = 29 and m = 65

So, it will be divisible by a-b. Therefore, we take (30-29) out of $$\dfrac{(30^{65}-29^{65})}{(30^{64}+29^{64})}=((30-29)*\dfrac{(30^{64}+30^{63}*29+....+29^{64})}{(30^{64}+29^{64})}$$
If you look closely at the first and last terms in the numerator, it is the same as the denominator. Thus Numerator is greater than the denominator.

It is greater than 1. Hence option D.

Hi,
We didn't expand, we just wrote it to the power of 64.

Hi Namit Bhargava,
If n is natural number, $$a^n-b^n$$ is always divisible by a - b 
Therefore, $$30^{65}-29^{65}$$ is written as $$\left(30-29\right)\left(30^{64}+30^{63}\left(29\right)+...+30\left(29\right)^{63}+29^{64}\right)$$
Hope this helps!