Originally Posted by
ThePerfectHacker In this expansion we have terms of form: $\displaystyle {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}$
To have a term independent of $\displaystyle x$ we require that the $\displaystyle x$ in numerator cancels out the $\displaystyle x$ in denominator. The exponent of $\displaystyle x$ in numberator is $\displaystyle k$ and the exponent of $\displaystyle x$ in denominator is $\displaystyle 3(8-k)$. We want that $\displaystyle k = 3(8-k)$ so that they cancel out. Therefore, $\displaystyle k=6$ if we solve this equation.
Therefore the indepentent term is: $\displaystyle {8\choose 6} (-1)^k 5^6$