Homework Help: How to simplify cube root expression

1. May 15, 2017

Daniel McKinley

]1. The problem statement, all variables and given/known data
Solving the cubic equation x^3 + 6x = 20 by using formula gives
(10+ sqrt(108))^1/3 - (-10 + sqrt(108))^1/3
How do you show that this comes out exactly 2? No calculators allowed.

2. Relevant equations

3. The attempt at a solution
Tried cubing the expression and tried using the Difference of Cubes formula.

2. May 15, 2017

scottdave

You can plug in for x (a - b) and get (a - b)^3 + 6(a-b) = 20. Expand out to get (a^3 - 3(a^2)b + 3a(b^2) - b^3) + 6a - 6b = 20. Were you able to get that far? Where did you get stuck? I'm not familiar off-hand with Diff of cubes. I will look that one up.

3. May 15, 2017

ehild

If the result is that simple, there must be a trick:) Try to find the numbers a and b so as (a+b)3=10+√108 and (a-b)3=-10+√108.
Hint : first factorize 108.
.

4. May 15, 2017

epenguin

you say you've tried but haven't showed what you have tried.

Firstly by substituting x = 2 in your equation you see that it is a solution. So (x - 2) is a factor of the cubic, so dividing by this term gives you a quadratic from which you can get the other roots.

However your question was just arithmetic.

√(108) ? 108 has a lot of factors so you can express √108 more naturally in terms of smaller numbers.

For the rest, I tried two avenues that both seem to be working, because seeming to be giving lots of simplifications, but in the end they didn't work. I might be onto something else but maybe tomorrow if no one else gets it first.

Last edited: May 16, 2017
5. May 15, 2017

epenguin

Okay I think this is it.
Our expression reduces to

(10 + 6√3) - (-10 + 6√3)

You hope that each of the two terms are perfect cubes. Maybe you can show they have to be. Maybe they have to be for this to give you an exact integer at the end.

Perfect cubes of something involving √3

Now (a + b√3)3 = a3 + 3a2b√3 + 9ab2 + 3b3√b

So we have two conditions to satisfy

For the non-surd term a3 + 9ab2 = 10

And the first thing you try:

a = 1, b = 1 works!

I will leave you to complete this now, and to worry about whether this will always work later.

6. May 16, 2017

epenguin

However I think this has all been probably unnecessary.
That you may be doing this at the wrong stage in the cubic solution or some other confusion.
In the cubic solution you do get the difference between two cubes
X3 - Y3 = 0
where X, Y are linear in the unknown, or Y is a constant.
Then you can use the difference between two cubes formula - but it is obvious anyway - to get one solution X = Y .

When you have worked out something please COME BACK and tell us about the problem and solution because I hate doing this amount and then a problem is left hanging and Incompleted And I hate the student who does it!
(PS Well now I see you are a teacher not a student so you should understand this.)

Last edited: May 16, 2017