How do you solve #root(3) { 5.67}#?

1 Answer
Dec 30, 2017

#root(3)(5.67) ~~ 1.783176586#

Explanation:

You can calculate approximations to the cube root of a number using Newton's method:

Given a number #n# whose cube root you wish to find and an approximation #a_i# to #root(3)(n)#, a better approximation is:

#a_(i+1) = a_i - (a_i^3-n)/(3a_i^2) = (2a_i^3+n)/(3a_i^2)#

Repeat to get successively better approximations.

For our example, with #n=5.67#, let #a_0 = 2#

Then:

#a_0 = 2#

#a_1 = (2a_0^3+n)/(3a_0^2) = (2(color(blue)(2))^3+5.67)/(3(color(blue)(2))^2) = (16+5.67)/12 = 1.8058bar(3)#

#a_2 = (2a_1^3+n)/(3a_1^2) = (2(color(blue)(1.8058bar(3)))^3+5.67)/(3(color(blue)(1.8058bar(3)))^2) ~~ 1.783459658#

#a_3 = (2a_2^3+n)/(3a_2^2) = (2(color(blue)(1.783459658))^3+5.67)/(3(color(blue)(1.783459658))^2) ~~ 1.783176631#

#a_4 = (2a_3^3+n)/(3a_3^2) = (2(color(blue)(1.783176631))^3+5.67)/(3(color(blue)(1.783176631))^2) ~~ 1.783176586#