A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of (2^(1/3) + 12(3)^(1/3))^10 is:

T5 = 10C4 (2^(1/3))^6 (12(3)^(1/3))^4

T15 = 10C6 (2^(1/3))^4 (12(3)^(1/3))^6

T5/T15 = 10C4 (2^(1/3))^6 (12(3)^(1/3))^4 / 10C6 (2^(1/3))^4 (12(3)^(1/3))^6

= 10C4 / 10C6 * (2^(1/3))^2 / (12(3)^(1/3))^2

Since 10C4 = 10C6

= (2^(1/3))^2 / (12(3)^(1/3))^2

= 2^(2/3) / (144 * 3^(2/3))

= 2^(2/3) / (144 * 3^(2/3))

= 1 / 72 * (2/3)^(2/3)

= (2^(2/3) ) / (144 * 3^(2/3))

= 1/72 (2/3)^(2/3)

Let's calculate T5 and T15 separately:

T5 = 10C4 (2^(1/3))^6 (12(3)^(1/3))^4 = 210 * 4 * 12^4 * 3^(4/3) = 210 * 4 * 20736 * 3√(81) ≈ 210 * 4 * 20736 * 4.3267 = 7506549.76

T15 = 10C6 (2^(1/3))^4 (12(3)^(1/3))^6 = 210 * 2^(4/3) * 12^6 * 3^2 = 210 * 2.2894 * 2985984 * 9 = 1280236389.16

T5/T15 = 7506549.76 / 1280236389.16 ≈ 0.00586

4(36)^(1/3) : 1