What is the constant term of the expansion of \(\left(6x+\dfrac{1}{3x}\right)^6?\)
By the Binomial Theorem, the constant term is 220.
Constant term
C(6,3) (6x)^3 ( 1 / (3x) )^3 =
20 * (6)^3 * x^3 * ( 1 / 27) * ( 1/x^3) =
20 * 6^3 / 27 =
20 * 216 / 27 =
20 * 8 =
160
Verified here : https://www.wolframalpha.com/input/?i=%286x+%2B+1%2F+%283x%29+%29%5E6
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