1. ## Newton's Binomial Theorem

(1 + x)^(1/3) = 1 + (1/3)x - (1/9)x^2 + (5/81)x^3 - (10/243)x^4 + (22/729)x^5

Use this to get a quick numerical estimate of 70^(1/3).

I started by observing that 70 = 10*(1+6)

so... 70^(1/3) = (10*(1+6))^(1/3) = 10^(1/3) * (1+6)^(1/3)

I then used the binomial expansion of (1 + x)^(1/3), plugging in x = 6

I am getting an answer no where close to the true value of 70^(1/3)

2. Hello, thamathkid1729!

$(1 + x)^{\frac{1}{3}} \;=\; 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{729}x^5 + \hdots$

$\text{Use this to get a quick numerical estimate of }70^{\frac{1}{3}}$

First, we must get $70^{\frac{1}{3}}$ in the form $(1 + x)^{\frac{1}{3}}$
. . This requires some Olympic-level gymnastics.

$70^{\frac{1}{3}} \;=\;(64 + 6)^{\frac{1}{3}}\;=\;\bigg[64\left(1 + \frac{6}{64}\right)\bigg]^{\frac{1}{3}}$

. . . . $=\;64^{\frac{1}{3}}\left(1 + \frac{3}{32}\right)^{\frac{1}{3}} \;=\;4\left(1 + \frac{3}{32}\right)^{\frac{1}{3}}$

$70^{\frac{1}{3}} \;=\;4\left(1 + \frac{3}{32}\right)^{\frac{1}{3}}$

. . . . $=\;4\,\bigg[1 + \frac{1}{3}\left(\frac{3}{32}\right) - \frac{1}{9}\left(\frac{3}{32}\right)^2 + \frac{5}{81}\left(\frac{3}{32}\right)^3 - \frac{10}{243}\left(\frac{3}{32}\right)^4 + \hdots \bigg]$

. . . . $=\;4(1.030321121)$

$70^{\frac{1}{3}} \;=\;\boxed{4.121284485}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check:

. . $(4.121284485)^3 \;=\;69.9999567$ . . . close enough!

3. Thank you! Do you know why it was not working for me when I used 70^(1/3) = (10*(1+6))^(1/3)

4. Originally Posted by thamathkid1729
Thank you! Do you know why it was not working for me when I used 70^(1/3) = (10*(1+6))^(1/3)
When the Binomial series is infinite, we require that |x|<1 for convergence.