Question

Expand the following using binomial theorem.

${(x^3-1)}^{\frac{1}{3}}{x}^5$

Answer 1

Now,

$\begin{array}{c}{\left(x^3-1\right)}^{\frac{1}{3}}{x}^5\\ =x^5\left(x-\frac{1}{3}{x}^{-2}-\frac{1}{9}{x}^{-5}..................\right)\\ =x^6-\frac{1}{3}{x}^{-3}-\frac{1}{9}..........\end{array}$