How do I find the constant term in ((-x)^3+(-3/x))^24 ?
I'm back here is how you finish.
$\displaystyle
([(-x)]^3+(-3x^{-1}))^{24}=\sum_{k=0}^{24}\binom{24}{k}(-x^{3})^{24-k}(-3x^{-1})^{k}
$
$\displaystyle \sum_{k=0}^{24} \binom{24}{k}(-1)^{24-k}x^{72-3k}(-3)^{k}(x)^{-k}= \sum_{k=0}^{24} \binom{24}{k}(3)^{k}x^{72-4k} $
Sinc ewe want the constant term we want the exponent on the x to be zero.
solving the equation
$\displaystyle 72-4k=0 \iff k=18$
we can generate the constant term using k=18
$\displaystyle \binom{24}{18}(-x^3)^{24-18}(-3x^{-1})^{18}= \binom{24}{18}(-x^3)^6\left(\frac{-3}{x} \right)^{18}=\binom{24}{18}(3)^{18}=52125248137444$