We have x = a 1 / 3 {\displaystyle x=a^{1/3}} or x 3 = a {\displaystyle x^{3}=a} .
We take f ( x ) = x 3 − a {\displaystyle f(x)=x^{3}-a} . Since ξ {\displaystyle \xi } is the exact root, we have ξ 3 = a {\displaystyle \xi ^{3}=a} .
Substituting x n = ξ + ϵ n , x n + 1 = ξ + ϵ n + 1 {\displaystyle x_{n}=\xi +\epsilon _{n},x_{n+1}=\xi +\epsilon _{n+1}} and a = ξ 3 {\displaystyle a=\xi ^{3}} in the given method, we obtain on simplification
ϵ n + 1 = ( p + q + r − 1 ) ξ + ( p − 2 q − 5 r ) ϵ n + 1 ξ ( 3 q + 15 r ) ϵ n 2 − 1 ξ 2 ( 4 q + 35 r ) ϵ n 3 + ⋯ {\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
p + q + r = 1 {\displaystyle p+q+r=1} ,
p − 2 q − 5 r = 0 {\displaystyle p-2q-5r=0} ,
3 q + 15 r = 0 {\displaystyle 3q+15r=0}
which gives p = 5 / 9 , q = 5 / 9 , r = − 1 / 9 {\displaystyle p=5/9,q=5/9,r=-1/9} .
The error equation becomes ϵ n + 1 = 5 3 ξ 2 ϵ n 3 + O ( ϵ n 4 ) {\displaystyle \epsilon _{n+1}={\frac {5}{3\xi ^{2}}}\epsilon _{n}^{3}+O(\epsilon _{n}^{4})}
or 
.
. Since
is the exact root, we have 
.
and 
in the given method, we obtain on simplification
,
,
.
