2006-06-24 07:21:28 · 13 answers · asked by twinklestar1120 1 in Travel ➔ Mexico ➔ Cancun
It depends on you and what you want to do and or see. Ive Benn to both locations, in fact ill be in Cancun 7/09/06 for my honeymoon. Bora Bora is without question the better place to vacation to me. Its unbelievably beautifully(look just like the postcards). There are seven shades of blue ocean, very nice locals, great tours, great food, good beer,perfect weather. Although expensive and very Little to no nightlife it is a trip of a lifetime. I highly recommend this island or other Tahitian islands. I will return when I can afford it. I cruised Tahiti so I cant recommend any resorts. Have a great time wherever you go.
2006-06-26 13:40:47 · answer #1 · answered by pobox01 2 · 0⤊ 0⤋
Bora Bora
2006-06-24 13:20:36 · answer #2 · answered by Marsha 2 · 0⤊ 0⤋
I just got back from Cancun two days ago, and I'ld go again. But have never been to Bora Bora...
2006-06-27 16:53:57 · answer #3 · answered by villaland 2 · 0⤊ 0⤋
I hated cancun!!! lots of students drunk on cheap beer/tequila...., expensive, bit hotels without character. Really a tourist trap.
Bora Bora ought to be better.
2006-06-24 07:26:41 · answer #4 · answered by OneLilithHidesAnother 4 · 0⤊ 0⤋
Bora Bora is spectacular and idyllic. Cancun doesn't even rate in comparison.
2006-06-24 13:08:50 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Bora Bora.
Cancun is run down, and too many tourists...
2006-06-24 07:26:10 · answer #6 · answered by magnamamma 5 · 0⤊ 0⤋
Depends on your budget. If you have a lot, than go to Bora Bora
2006-06-28 12:22:40 · answer #7 · answered by beatricearuste@rogers.com 3 · 0⤊ 0⤋
depends on how much money you want to spend and how long you want to be on a plane!!!
botg are fabulous and beautiful!!
but bora bora is obviously the more expensive of the two!
2006-06-27 04:46:40 · answer #8 · answered by Sarah R 2 · 0⤊ 0⤋
bora bora
2006-06-24 07:24:15 · answer #9 · answered by tapdiva2003 3 · 0⤊ 0⤋
cancun
2006-06-24 07:28:35 · answer #10 · answered by Anonymous · 0⤊ 0⤋