# Cube root of 5

• May 6th 2011, 03:14 AM
mathguy80
Cube root of 5
Hi Guys,

Need a little hint on this one.

Expand $(1 + x)^{\frac{1}{3}}$ in ascending powers of $x$ as far as the term in $x^3$, simplifying the terms as much as possible. By substituting 0.08 for $x$ in your result obtain an approximate value of the cube root of 5, upto four decimal places.

I expanded the problem as a binomial series,

$1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \dfrac{5x^3}{81}$

The rest of the question is confusing me. How do I break up 5 to include 0.08?

The only approach I could think of was,

$
\sqrt[3]{5} = \sqrt[3]{8\left(1 - \frac{3}{8}\right)} = 2\sqrt[3]{1 - \frac{3}{8}}
$

Then using x = -3/8, to put in the above equation.

This gives me 1.7122 which is incorrect, and also does not use the suggested 0.08. Any ideas where I am going wrong?

• May 6th 2011, 03:21 AM
alexmahone
$(1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{25}$

Can you proceed?
• May 6th 2011, 03:22 AM
Prove It
Surely if you're evaluating the cube root of 5, you need to let x = 4...
• May 6th 2011, 03:39 AM
mathguy80
Quote:

Originally Posted by alexmahone
$(1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{5}$

Can you proceed?

I am missing something here, Wouldn't that be $3/\sqrt[3]{25}$? How does that lead to $\sqrt[3]{5}$?
• May 6th 2011, 03:51 AM
alexmahone
Quote:

Originally Posted by mathguy80

I am missing something here, Wouldn't that be $3/\sqrt[3]{25}$? How does that lead to $\sqrt[3]{5}$?

Of course. I'm sorry, that was a typo.

Anyway, $3/\sqrt[3]{25}=3/(\sqrt[3]{5})^2$
• May 6th 2011, 04:09 AM
mathguy80
Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?
• May 6th 2011, 04:12 AM
mathguy80
• May 6th 2011, 04:50 AM
Quote:

Originally Posted by mathguy80
Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?

$\sqrt[3]{1.08}=\sqrt[3]{\frac{108}{100}}=\sqrt[3]{\frac{(4)27}{(4)25}}$

$\Rightarrow\sqrt[3]{5}=\sqrt[3]{\frac{500}{100}}=\sqrt[3]{\left(\frac{108}{108}\right)\frac{500}{100}}$

$=\sqrt[3]{\frac{500}{108}\left(\frac{108}{100}\right)}}$

$=\sqrt[3]{\frac{500}{108}}\sqrt[3]{1+0.08}=\sqrt[3]{\frac{(4)125}{(4)27}}\sqrt[3]{1+0.08}$

$=\sqrt[3]{\frac{125}{27}}\sqrt[3]{1+0.08}$
• May 6th 2011, 06:29 AM
Prove It
Quote:

Originally Posted by mathguy80

Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...
• May 6th 2011, 06:32 AM
alexmahone
Quote:

Originally Posted by Prove It
Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...

But the binomial series (when the index is a fraction) converges only if |x|<1.
• May 6th 2011, 06:53 AM
Prove It
Quote:

Originally Posted by alexmahone
But the binomial series (when the index is a fraction) converges only if |x|<1.

Point taken. I'm sure, however, that a series can be found which is centred somewhere near x = 4 so that you can evaluate it at x = 4.
• May 6th 2011, 08:28 AM
mathguy80
Nice! This is why I love this site! Thanks @Archie Meade.
• May 6th 2011, 08:32 AM
mathguy80
Interesting point @Prove it. The |x| < 1 for the convergent binomial series has tripped me up as well. Don't know enough Math to know if there is such a series.. Thanks for all the help today, guys, much appreciated.