We have ${\displaystyle x=a^{1/3}}$ or ${\displaystyle x^{3}=a}$.

We take ${\displaystyle f(x)=x^{3}-a}$. Since${\displaystyle \xi }$ is the exact root, we have ${\displaystyle \xi ^{3}=a}$.

Substituting ${\displaystyle x_{n}=\xi +\epsilon _{n},x_{n+1}=\xi +\epsilon _{n+1}}$ and ${\displaystyle a=\xi ^{3}}$ in the given method, we obtain on simplification

${\displaystyle \epsilon _{n+1}=(p+q+r-1)\xi +(p-2q-5r)\epsilon _{n}+{\frac {1}{\xi }}(3q+15r)\epsilon _{n}^{2}-{\frac {1}{\xi ^{2}}}(4q+35r)\epsilon _{n}^{3}+\cdots }$

For the method to be of order 3, we have

${\displaystyle p+q+r=1}$,

${\displaystyle p-2q-5r=0}$,

${\displaystyle 3q+15r=0}$

which gives ${\displaystyle p=5/9,q=5/9,r=-1/9}$.

The error equation becomes ${\displaystyle \epsilon _{n+1}={\frac {5}{3\xi ^{2}}}\epsilon _{n}^{3}+O(\epsilon _{n}^{4})}$