Analyzing convergence of finding roots for $f(x) = x^3 - a$ using Newton's method

Question

Given $f(x) = x^3 - a$, we wish to find $f(x^*)=0$ as a way to calculate the cube root of a number. This can be done using Newton's method. So we have $x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$. The problem asks to show that this in fact quadratically converges as Newton's method should. So essentially, we want to show $|x_{k+1}-x^*| \leq M|x_k-x^*|^2$. However, I have had a hard time isolating $x_k$ in this case: $$x_{k+1}-x^* = x_k - x^*-\frac{x_k^3-a}{3x_k^2}$$ I have played around with this for a while but I have not had any luck. Any help is appreciated, thank you.

Answer 1

Replace $a$ with $(x^*)^3$ and apply binomial theorems to get $$ x_{k+1}-x^* = (x_k-x^*)\,\frac{3x_k^2-(x_k^2+x_kx^*+(x^*)^2)}{3x_k^2} =(x_k-x^*)^2\,\frac{2x_k+x^*}{3x_k^2}. $$