# Thread: Substitute suitable value of y to find 5^{1/3}

1. ## Substitute suitable value of y to find 5^{1/3}

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

2. ## Re: Substitute suitable value of y to find 5^{1/3}

$\displaystyle \sqrt[3]{5} \approx 1.709$ and $\displaystyle \frac{149}{225} \approx 1.510$.

They are almost equal.

3. ## Re: Substitute suitable value of y to find 5^{1/3}

Originally Posted by AgentSmith
$\displaystyle \sqrt[3]{5} \approx 1.709$ and $\displaystyle \frac{149}{225} \approx 1.510$.

They are almost equal.
In my opinion, they differ very much because if $\displaystyle y=\frac{1}{5}$ is used instead, the answer is $\displaystyle 1.72$ which is very much closer to the exact answer. The problem over here is that I don't understand why such a huge difference was generated when it seemed all reasonable to use $\displaystyle y=-\frac{4}{15}$

4. ## Re: Substitute suitable value of y to find 5^{1/3}

Originally Posted by Punch
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.
Because $\displaystyle 1/5$ is closer to $\displaystyle 0$ than $\displaystyle -4/15$ .