[General Battuta]

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.

Of course we did pure algebraic manipulations as well, both in pure math and in topics like special relativity. But numerical calculations come up all the time in physics and other mathematical applications too. It's not a question of whether or not you're allowed to do them, sometimes you need to.

[Locutus of Borg]

lol no

[Mika]

I don't know to which question you answered, but here goes.

Why would you want to cube the nasty looking cubic root side? That's tedious work, I think your original approach was better. In fact, we would like to show that under that cubic root, there indeed is a cube of something. Why? Because that makes it rather easy to calculate and the cubic roots are cancelled.

So in principle, the effect should be [(a - b)³]^1/3 = a - b

How could you know that 20-14*sqrt(2) is a cube of something? Well, you are actually cheated by the looks of it, since using numbers allows combining terms in a way that it is very non-intuitive to get back in to the expanded cubic form of a³ - 3a²b + 3ab² - b³. Which is the reason I still heartily recommend staying the hell away from the numbers until last possible moment and said this is a rather arbitrary and stupid kind of assignment.

So now we are basically saying that a³ - 3a²b + 3ab² - b³ = 20 - 14*sqrt(2)

That's fine, but not very informative. We have two unknowns a and b and only one equation. To solve the equation exactly, we would need two independent equations. Where do we get the second one?
Well, there is a bit of a hint in the numerical form, so we can actually write:

(1) a³ + 3ab² = 20
(2) -3a²b - b³ = -14*sqrt(2)

Now, as I said I wont solve the problem myself, so spoon-feeding the answer stops here. Explain me why did I write these two equations; I wont continue any further until I get reasonable feedback from you. I'm positive that you will get it, but it might take some time. This is where the assignment actually becomes ingenious.

[Locutus of Borg]

Well since b is the radical then the terms that have it to an even power will result in an integer

the terms that have the radical to an odd power will have it as a radical still

Those two equations make a system. In order to solve it we need to make them both equal the same thing. (Adding or subtracting the equations gets us back where we started :@)

[Mika]

Your deduction is basically correct.

There are a couple of manipulation steps for the two equations.

Let's mark b=C*sqrt(2), where C is a constant. Why? Because we know that the sqrt(2) has to be a participant of the equation (2), but we don't know for sure if it had a multiplier in front of it.

The equations turn to:
(1) a³ + 6aC² = 20
(2) -3a²C - 2C³ = -14

And to make it a little bit more clean, multiply (2) with -1.

(1) a³ + 6aC² = 20
(2) 3a²C + 2C³ = 14

Here, there are two options for continuation, either you might guess the solution for both equations, or, there is the analytical solution for them, but from this point on it gets rather complex.

I will not do further steps today as I need to go to sleep, if you really want to see how complex it can get, solve a as C's function. Let's see what's happened here tomorrow then.

[Locutus of Borg]

Turns out I copied the question wrong. It really asks for x^3 - 6x

if the contents under the first radical (the plus one) is a and the contents under the second radical (the minus one) is b then

x = a^1/3 + b^1/3
x^3 = a + a^2/3 b^1/3 + a^1/3 b^2/3 + b

a+b = 40
6x = 6a^1/3 - 6b ^ 1/3

x^3 - 6x = 40 + 3a^2/3 b^1/3 + 3a^1/3b^2/3 - 6a^1/3 - 6b^1/3

factor out (3)(A^1/3 + b^1/3) from the right

x^3 - 6x = 40 +3 (a^1/3 + b^1/3)(a^1/3 b^1/3 - 2)

a^1/3 + b^1/3 = x so

x^3 - 6x = 40 + 3x ((ab^1/3 - 2)

ab^1/3 = (400 - 391)^1/3 = 2

2 - 2 = 0!

x^3 - 6x = 40 + 0
x^3 - 6x = 40

[Mika]

Yeah, I was kinda thinking that there has to be something wrong with the assignment, as solving the earlier one by the book would have required the formula for solving a third order polynomial... Perhaps even complex numbers.

Next time be more careful.

[Locutus of Borg]

Thanks for your help anyway