# How to simplify?

• Aug 17th 2013, 09:43 PM
How to simplify?
Question directions simply state to simplify: (hehe, simply simplify)
(20+14sqrt(2))^(1/3) + (20-14sqrt(2))^(1/3)

WolframAlpha solution says to express 20 + 14√(2) as a cube to equal 8 + 12√(2) + 6((√2))^2+(√(2))^3. Can someone explain how these numbers were crunched up? Is it to do with (a^3+b^3)=(a+b)(a^2-ab+b^2)?
• Aug 17th 2013, 10:24 PM
chiro
Re: How to simplify?

How do you want to simplify this? If its in terms of common factors multiplication wise then I don't think you can do that.
• Aug 17th 2013, 11:19 PM
agentmulder
Re: How to simplify?
As far as i know this kind of thing relies a lot on luck. We can try

$\sqrt[3]{20 + 14 \sqrt{2}} = a + b \sqrt{2}$

cube both sides to get

$20 + 14 \sqrt{2} = a^3 + 6ab^2 + (3a^2b + 2b^3) \sqrt{2} \ \ \ \ \$ (1)

Now we see

$a^3 + 6ab^2 = 20$

$3a^2b + 2b^3 = 14$

Luckily , we find with very little effort, a = 2 , b = 1 works so

$\sqrt[3]{20 + 14 \sqrt{2}} = a + b \sqrt{2} = 2 + 1 \sqrt{2}$

Repeating the procedure using the right hand side of \$(1) , you don't have to start from scratch , we find the other cube root must satisfy

$20 - 14 \sqrt{2} = a^3 + 6ab^2 + (3a^2b + 2b^3) \sqrt{2}$

$a^3 + 6ab^2 = 20$

$3a^2b + 2b^3 = -14$

And again with little effort we find a = 2 , b = -1 works so

$\sqrt[3]{20 - 14 \sqrt{2}} = a + b \sqrt{2} = 2 + (-1) \sqrt{2}$

So the sum of your cube roots is actually the very humble integer 4.

$\sqrt[3]{20 + 14 \sqrt{2}} + \sqrt[3]{20 - 14 \sqrt{2}} = 2 + \sqrt{2} + 2 - \sqrt{2} = 4$

He heh...

:D
• Aug 18th 2013, 12:04 AM
Re: How to simplify?
Ahhh! I knew it had to be a guess and check method! There is no systematic way other than plugging away at numbers? There will be a huge time constraint when I have to deal with this during my exam :( Oh well, thank you very much! :D
• Aug 19th 2013, 07:19 AM
Idea
Re: How to simplify?
$x=\left(a+b\sqrt{d}\right)^{1/3}+\left(a-b\sqrt{d}\right)^{1/3}$

is the unique real solution of the cubic equation

$x^3-3p x - 2a =0$

where

$p=\left(a^2-b^2 d\right)^{1/3}$
• Aug 19th 2013, 09:16 PM
Re: How to simplify?
Quote:

Originally Posted by Idea
$x=\left(a+b\sqrt{d}\right)^{1/3}+\left(a-b\sqrt{d}\right)^{1/3}$

is the unique real solution of the cubic equation

$x^3-3p x - 2a =0$

where

$p=\left(a^2-b^2 d\right)^{1/3}$

How in the world... that is so cool! Where did you derive these formulas from? Are there similar forms for $x=\left(a+b\sqrt{d}\right)^{1/2}+\left(a-b\sqrt{d}\right)^{1/2}$ ?
• Aug 20th 2013, 08:48 PM
Idea
Re: How to simplify?
Quote:

How in the world... that is so cool! Where did you derive these formulas from? Are there similar forms for $x=\left(a+b\sqrt{d}\right)^{1/2}+\left(a-b\sqrt{d}\right)^{1/2}$ ?
$x=\left(a+b\sqrt{d}\right)^{1/2}+\left(a-b\sqrt{d}\right)^{1/2}$
$x^2=2a+2\sqrt{a^2-b^2d}$
$\left(17+12\sqrt{2}\right)^{1/2}+\left(17-12\sqrt{2}\right)^{1/2} = 6$