Correct Answer - a
We have,
`(root(8)(5)+ root(6)(2))^(100) = sum_(r=0)^(100) ""^(100)C_(r) (root(8)(5))^(100 - r) (root(6)(2))^(r)`
`rArr (root(8)(5)+ root(6)(2))^(100) = sum_(r=0)^(100) ""^(100)C_(r) 5(100 - r)/(8) (r)/(2^(6)) sum_(r=0)^(100) T_(r + 1), `
where `T_(r+1) =""^(100)C_(r) 5 (100 - r)/(8) (r)/(2^(6)`
Clearly, `T_(r+ 1) ` will be an integer if `(100 - r)/(8) and (r)/(6)` are integers.
This is possible when 100 - r is a multiple of 8 and r is a multiple of 6
`rArr 100 - r = 0 , 8 16 ,...., 96 and r = 0, 6, 12,...96`
`rArr r = 4, 12, 20,..., 100 and r = 0, 6, 12,..., 96`
`rArr r = 12 , 36, 60, 84`
Hence, there are 4 rational terms and 101-4 = 97 irrational terms.
