Given the equation $\displaystyle (1+3y)^{\frac{1}{3}} \approx 1+y-y^2$ is valid for $\displaystyle |y|<\frac{1}{3}$, by substituting a suitable value for $\displaystyle y$, find an approximate value of $\displaystyle 5^{\frac{1}{3}} $.

I found the answer by substituting $\displaystyle y=\frac{1}{5}$ and the answer is $\displaystyle 1\frac{21}{29}$

But I dont understand why, when I used $\displaystyle y=-\frac{4}{15}$, i dont get the answer.

Here are my workings.

When $\displaystyle y=-\frac{4}{15},$

$\displaystyle (\frac{1}{5})^{\frac{1}{3}}\approx 1+\frac{-4}{15}-\frac{16}{225}=\frac{149}{225}$

Therefore, $\displaystyle 5^{\frac{1}{3}}=\frac{225}{149} $ which is wrong