A constant terms comes out when a^n-k * b^k is a constant, and does not involve any variables. In your case, the binomial expansion has only one variable, x. That means, in (3x^6)^n - k * (5/x)^k does not contain any x. For it to happen, when n = 21, you need to equate the power of x to zero, since x^0 = 1, is a constant, and other powers yield terms with x, and so, not a constant term.

(x^6)^21 - k * (1/x)^k = x^0

(x^6)^21 - k * (x^-1)^k = x^0

(x^6)^21 - k * x^-k = x^0

x^(6(21-k) - k) = x^0

Equating coefficients, you get k = 18, which is what he used to get the constant term above.

Salahuddin

Maths online