Finding parameter so Newton Method converges cubically Consider the function $\phi(x)=x^b(x^2-a)$. Find a value for $b\neq 0$ independent of $a$ so that the Newton Interative Method for $\phi$ locally converges cubically to $\sqrt{a}$. I know that for local cubic converges of the Newton method I need to have $\phi''(x^*)=0$ where $x^*$ is the fix point we are looking for. When I calculate the second derivative I get this; $$\phi''(x)=b^2x^b+3bx^b+2x^b-b^2ax^{b-2}+bax^{b-2}$$ What am I missing here?

We set $\phi''(\sqrt{a}\,)=0$ using these steps: \begin{align*} \phi(x)&=x^{b+2}-a x^b\\\ \phi'(x)&=(b+2)\,x^{b+1}-ab\,x^{b-1}\\\ \phi''(x)&=(b+1)(b+2)\,x^b-ab(b-1)\,x^{b-2}\\\ \phi''(\sqrt{a}\,)&=(b+1)(b+2)\,a^{b/2}-ab(b-1)\,a^{b/2-1}\overset{\text{set}}{=}0. \end{align*} The $a^{b/2}$ factor is never zero, so we eliminate it to obtain $$0=(b+1)(b+2)-ab(b-1)a^{-1}=b^2+3b+2-b^2+b=4b+2,$$ with solution $$b=-\frac12.$$