Determine the constant term in.
idk how to solve it
each term in a binomial expansion is of the form ...
$\displaystyle \binom{n}{k} a^{n-k} b^k$
In this case, $\displaystyle n = 21$ ... the value of $\displaystyle k$ will be such that the variable factors cancel.
$\displaystyle \binom{21}{18} (3x^6)^3\left(\frac{5}{x}\right)^{18}$
your book is not the sole source of knowledge. seek and ye shall find ...
Binomial Theorem (part 1) | Precalculus | Khan Academy
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