Find the power series representation centered at $x=0$ of the following function. Give the interval of convergence for the resulting series. 
$f\left(x\right)=\frac{1}{1-x^3}$

• A. $\sum_{n=0}^{\infty}x^{3n}$; converges on (-∞,∞)
• B. ${\sum }_{n=0}^{\infty }{x}^{3n}$; converges on (-1,1)
• C. ${\sum }_{n=0}^{\infty }\frac{1}{x^{3n}}$; converges on (-1,1)
• D. ${\sum }_{n=0}^{\infty }\frac{1}{x^{3n}}$; diverges everywhere

Answer: (B)

Verified step by step guidance

1. Step 1: Recognize that the given function f(x) = 1 / (1 - x^3) resembles the geometric series formula, which is 1 / (1 - r) = ∑_{n=0}^{∞} r^n, valid for |r| < 1. Here, r = x^3.
2. Step 2: Rewrite the function as a geometric series using the formula. Substitute r = x^3 into the series expansion: f(x) = ∑_{n=0}^{∞} (x^3)^n = ∑_{n=0}^{∞} x^{3n}.
3. Step 3: Determine the interval of convergence for the series. The geometric series converges when |r| < 1. Since r = x^3, we require |x^3| < 1, which simplifies to |x| < 1.
4. Step 4: Conclude that the series representation of f(x) is ∑_{n=0}^{∞} x^{3n}, and the interval of convergence is (-1, 1).
5. Step 5: Verify the endpoints of the interval of convergence. At x = -1 and x = 1, the series ∑_{n=0}^{∞} x^{3n} does not converge because the terms do not approach zero. Therefore, the series converges only within (-1, 1).