Answer to Question #242824 in Algebra for Anand

Question

Find the approximate value of ³√0.99 up to 4 decimal places.

Answer 1

Let b³=0.99. Also, let us assume that a=1 an ϵ be a small correction factor such that b=a-ϵ.

Now, cubing both sides:

b³=a³-3a²ϵ+3aϵ²-ϵ³

$0.99=1-3ϵ+3ϵ^2-ϵ^3$

Assuming that ϵ³<<ϵ²<<ϵ :

$0.99=1-3\varepsilon$

$\varepsilon=(1-099)/3=0.0033$

$\sqrt[3]{0.99}=1-0.0033=0.9967$