1. **State the problem:** Find the power series representation for the function $$f(x) = x^8 \tan^{-1}(x^3)$$ and determine its radius of convergence $$R$$. 2. **Recall the power series for arctangent:** The function $$\tan^{-1}(z)$$ has the power series expansion $$ \tan^{-1}(z) = \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{2n+1}, \quad |z| < 1. $$ 3. **Substitute $$z = x^3$$:** $$ \tan^{-1}(x^3) = \sum_{n=0}^\infty (-1)^n \frac{(x^3)^{2n+1}}{2n+1} = \sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{2n+1}. $$ 4. **Multiply by $$x^8$$:** $$ f(x) = x^8 \tan^{-1}(x^3) = x^8 \sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{2n+1} = \sum_{n=0}^\infty (-1)^n \frac{x^{6n+11}}{2n+1}. $$ 5. **Final power series representation:** $$ f(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{6n+11}}{2n+1}. $$ 6. **Determine the radius of convergence:** Since the original series for $$\tan^{-1}(z)$$ converges for $$|z| < 1$$, here $$z = x^3$$, so convergence requires $$ |x^3| < 1 \implies |x| < 1. $$ Thus, the radius of convergence is $$ R = 1. $$