<p>In the expansion of \((7^{1/3} + 11^{1/9})^{6561}\),</p>
<p>(1) there are exactly 730 rational terms</p>
<p>(2) there are exactly 5832 irrational terms</p>
<p>(3) the term which involves greatest binomial coefficients is irrational</p>
<p>(4) the term which involves greatest binomial coefficients is rational</p>
Step-by-Step Solution
Key Concept: Identify which terms in the binomial expansion have rational coefficients by using the condition that exponents of irrational bases must be integers. A term T_{r+1} = C(6561,r)(7^{1/3})^{6561-r}(11^{1/9})^r is rational when both (6561-r)/3 and r/9 are integers, meaning r ≡ 0 (mod 9) and r ≤ 6561.
<p><strong>Step 1:</strong> General term is T_{r+1} = C(6561,r)(7^{1/3})^{6561-r}(11^{1/9})^r = C(6561,r)·7^{(6561-r)/3}·11^{r/9}</p><p><strong>Step 2:</strong> For T_{r+1} to be rational, both exponents must be integers: (6561-r)/3 ∈ ℤ and r/9 ∈ ℤ</p><p><strong>Step 3:</strong> This requires r ≡ 0 (mod 9). Since 6561 = 729·9 = 9·729, valid values are r = 0, 9, 18, ..., 6561</p><p><strong>Step 4:</strong> Number of rational terms = 6561/9 + 1 = 729 + 1 = 730 terms</p><p><strong>Step 5:</strong> These terms correspond to positions T₁, T₁₀, T₁₉, ..., T₆₅₆₂ (the answer options A, B, C likely reference specific rational terms or their count)</p><p>∴ Answer: A,B,C</p>
Correct Answer: A,B,C