[Locutus of Borg]

Why

(20 - 14 * 2^(1/2))^(1/3) = 2 - (2)^1/2

[perihelion]

Erm, (P)lease (E)xcuse (M)y (D)ear (A)unt (S)ally?

[General Battuta]

(20-14*root2)^1/3  = 2 - root2

root2 = 1.4142

(20 - 14 * 1.4142) ^ 1/3 = 2 - 1.4142

(20 - 19.7988) ^ 1/3 = .5858

(.2012) ^ 1/3 = .5858

.59 = .5858, the error's just introduced by rounding earlier

[Locutus of Borg]

How do I arrive at 2 - (2)^1/2 though

I need to write a proof starting with the first term.

[Mika]

Battuta, I think any idiot can do that substitution and doesn't help OP to learn about symbolic calculations.

The easiest way to derive that expression is simply to cube both sides of the equation.

20-14*sqrt(2)=(2-sqrt(2))³

Now open the right hand side and remember that (a-b)³ = a³ - 3a²b + 3ab² - b³

You can then collect the terms. More than that I'm not gonna say, and I may have already said too much.

[General Battuta]

Battuta, I think any idiot can do that substitution and doesn't help OP to learn about symbolic calculations.

I'm well aware of that, but the OP didn't make it clear that he wanted anything more than a simple arithmetic resolution, nor did it suggest anything about learning about symbolic calculations. As far as we knew this was an order of operations confusion.

[Mika]

Or, if you didn't know the polynomial expansion of (a - b)³, you might need to derive it there. But that should be easy.

ADDENDUM:

Let's try to put it down one more time: either derive the expansion of (a-b)³ , or substitute the numbers to (a-b)³ and simplify. I recommend deriving the expression, though, since if you get a job like mine, more often than not it is the derivations of the equations that are more important than just plunging in the numbers.

[Locutus of Borg]

My problem is that I figured out that the expression equals 2 - (2)^1/2 on a calculator. I need to explain how I got there from (20 - 14 * 2^(1/2))^(1/3)

This is all part of a larger question but once this is solved then the rest is simple.

(and yes I know that (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

[z64555]

I'd suggest following Mika's method. Don't plug anything into the calculator, nor try to evaluate the sqrt(2) until the very end.

You should essentially come out with an equation of sums on one or both sides of the equal sign.

Btw, I ended up with 20 = 20 

[Locutus of Borg]

Here's the entire question

if x = (20 - 14 * 2^(1/2))^(1/3) + (20 + 14 * 2^(1/2))^(1/3)

What is x^3

I know the answer is 64 but I can't PROVE it.

[Qent]

I cubed both sides (of the original question) and expanded the right side. It seems kinda random, but I just figured exponents would be easier to work with than roots.

[Locutus of Borg]

But then you're still left with fractional exponents.

[General Battuta]

Fractional exponents are fine, you can always split them into a power and a root if it's easier to look at for you.

[Davros]

I figured out that the expression equals 2 - (2)^1/2 on a calculator. I need to explain how I got there

Thats easy just be truthful write as your answer "I typed it into a calculator"

Davros leaves the thread the audience are stunned into silence by his totally awesome mathematics prowess

[Mika]

Disclaimer: I don't do home assignments for people, but I might help in solving them. Work needs to be done by the OP himself.

That being said, due to the perceived stupidity/geniosity (that's right) of this assignment, I might help you to do this, but first of all you need to show me you have really done something yourself.
(1) Post where you are at the moment, and what you have tried and why.

(2) Then tell me your age and what you know about counting with abstract letters instead of numbers.

I'm not trying to be rude, (1) is needed to convince me that you have indeed done something yourself. (2) is needed because I suspect this assignment is beyond the capabilities of the average person of your age.

[General Battuta]

We did this kind of stuff as college freshmen (or heck high school junior/seniors) all the time. Maybe even earlier.

[Locutus of Borg]

I said that the 20 + 14sqroot(2) = a
and 20 - 14sqroot(2) = b

a^3 + 3a^2b + 3ab^2 + b^3

a^3 + b^3 = 40

3a^2b = 160 + 112sqroot(2)

that's where I am right now. I'm not working on it atm.

[Mika]

A couple of additional questions, explain me why would you want to cube immediately the equation

x = (20-14*sqrt(2))^(1/3)+(20+14*sqrt(2))^(1/3)?

Also, please answer (2).

I know this is kind of grilling, but I need to make sure you have at least done something yourself.

We did this kind of stuff as college freshmen (or heck high school junior/seniors) all the time. Maybe even earlier.

If you were really allowed to use numbers in calculations like this during high school, I'm indeed pretty much speechless.

[Locutus of Borg]

I'm cubing it because I need find x^3 - 6

2: I  know a little calculus, so I am old enough to be able to do this 

[Mika]

The question is, how well can you calculate with polynomials that consist only of letters? That is, If you want more formal resolution of the problem, you need to be able to solve a set of two equations that only consist of letters. I initially thought you were around 14, would that be anywhere near correct?