Hello, its a simple problem but I am not able to spot my fault. Please assist.

Q) what is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be 6?

Solution:
Let A be the event that the number cube rolls a 6.
P(A) = 1/6
P(not A) = 5/6

As it is asked, what is probability of at least one of the rolls to be 6,
we can quickly calculate the probability of no rolls to be 6, which is equal to (5/6)*(5/6)*(5/6) = 125/216.
Hence, probability of at least one of the rolls to be 6 = 1 - 125/216 = 91/216

My doubt:
If I think of calculating the probability in direct way, please tell me where I am making mistake.

We are asked : prob. of at least one of the rolls be a 6 == it means, one roll could be a 6 OR two rolls could be a 6 OR all 3 rolls could be a 6

therefore, ans = P(one roll be a 6) + P(two rolls be a 6) + P(all 3 rolls be a 6)
ans = ((1/6)*(5/6)*(5/6)) + ((1/6)*(1/6)*(5/6)) + ((1/6)*(1/6)*(1/6))
ans = 31/216

Please help me correct my discrepancy between getting the answer as 91/216 and 31/216.
Thanks!