Comparing

x$1=3$

and

x$2=2\mathrm{.}91666667,$

we see that they do not exactly match to the required eight decimal places. Therefore, we must apply Newton's method again with

f$($x$)=$ x$472$

and

f$($x$)=4$x$3$

to find the third approximation

x$3.$

Doing so and rounding to eight decimal places gives the following result.

x$3=$ x$2$

f$($x$2)$f$($x$2)$

$$

$=$