Rational Term

Question

Within the expansion of (\sqrt{3} + \sqrt[3]{2})^{6}, the number of terms that turn out to be rational, free of any surd, is counted to be which value?

RankGuru · Accepted Answer

A term is rational only when every fractional exponent simplifies to an integer, so locating rational terms means imposing simultaneous integrality conditions on the general term, a precise JEE Advanced surd analysis. The general term is T_{r+1} = C(6, r) (3^{1/2})^{6-r} (2^{1/3})^{r} = C(6, r) 3^{(6-r)/2} 2^{r/3}. For rationality we need (6 - r)/2 to be an integer and r/3 to be an integer simultaneously, with r ranging from 0 to 6. The condition r/3 ∈ Z forces r ∈ {0, 3, 6}. Among these we test (6 - r)/2: for r = 0 it is 3 (integer, valid); for r = 3 it is 3/2 (not an integer, rejected); for r = 6 it is 0 (integer, valid). Thus exactly two values, r = 0 and r = 6, satisfy both conditions, giving two rational terms. At r = 0 the term is C(6,0) 3^3 = 27, and at r = 6 it is C(6,6) 2^2 = 4, both free of any surd. Hence the count is 2. Option 1 undercounts by dropping one of the two valid values of r. Option 3 wrongly accepts r = 3, which leaves the half-integer exponent 3/2 on 3 and so retains a square-root factor. Option 4 grossly overestimates. Plausibility check: r/3 ∈ Z gives candidates {0, 3, 6}, and the extra parity requirement 2 | (6 - r) discards only r = 3, leaving exactly two rational terms.

Rational Term

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