\(-15a^{4}b^{2}\)
\(15a^{4}b^{2}\)
\(-15a^{3}b^{3}\)
\(15a^{3}b^{3}\)
Correct answer is B
\((a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...\)
Third term = \(^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)\)
= \(15a^{4}b^{2}\)
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