# Homework Help: Evaluate expression W/O Calculator

1. Jan 31, 2012

### kscplay

1. The problem statement, all variables and given/known data

Find the exact value of the following expression.

$\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}$

2. Relevant equations

3. The attempt at a solution

Don't know where to start. I guess we can factor out a 2$1/3$ but I don't see how that would help. Any clues? Perhaps some factoring method/identity can be used? Thanks.

2. Feb 1, 2012

### Curious3141

Let $x = \sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}$

Find $x^3$.

3. Feb 1, 2012

### SammyS

Staff Emeritus
Maybe there are values for a & b such that
$\displaystyle \left(a+b\sqrt{3}\right)^3=10+6\sqrt{3}$​
and
$\displaystyle \left(a-b\sqrt{3}\right)^3=10-6\sqrt{3}$​
...

or something like that.

4. Feb 1, 2012

Ingenious!

ehild

5. Feb 1, 2012

### Curious3141

Thanks, but not really!

6. Feb 1, 2012

### ehild

Why not? You get an easily factorizable equation of form x^3+ax+b=0.

ehild

7. Feb 1, 2012

### SammyS

Staff Emeritus
The answer is: a=1, b=1 .

$\displaystyle \left(1\pm\sqrt{3}\right)^3=10\pm6\sqrt{3}$

8. Feb 1, 2012

### Curious3141

No, I meant it's not really ingenious, I found it easy to see, and I'm sure it's quite easy for others to see, too.

Familiarity with the expansion of the binomial form (a+b)3 will indicate there are square terms which might help with the square root surd (in the event, it does not, but that's what led me to think of it). Combined with the fact that the two factors are conjugate surds which can be multiplied together to give an integer (a perfect cube!), it's quite easy to see that this is a quick and painless way.

Yes, you get a cubic that can be easily solved with the rational root theorem. The other solutions are complex, so the only real (integral) root is the answer.

9. Feb 1, 2012

### kscplay

Okay, I get the following for x3:

20-6$\sqrt[3]{10+6\sqrt{3}}$-6$\sqrt[3]{10-6\sqrt{3}}$

I still don't see it. Is my x3 correct? Thanks.

10. Feb 2, 2012

### kscplay

Could you please elaborate on that? Thanks.

11. Feb 2, 2012

### ehild

Notice that x3=20-6x.

Or choose SammyS's hint...

ehild

12. Feb 2, 2012

### ehild

Rewrite the equation as (x3-8)+6(x-2)=0. You can factor out x-2.

but SammyS' method is easier: $$\displaystyle \left(1\pm\sqrt{3}\right)^3=10\pm6\sqrt{3}$$
He is the Master of Square Roots

ehild

Last edited: Feb 2, 2012
13. Feb 2, 2012

### kscplay

Thanks, I get it now :) My final answer is 2. I just factored because I didn't really understand how SammyS solved for a & b.

14. Feb 2, 2012

### ehild

He has got magical eyes. He just looks at an expression with roots and sees how to simplify it. It comes with long practice, I guess.

But it is worth to remember that the powers of (b±√a) are of the form A+B√a.

(b±√a)3=b3±3b2√a+3ab±a√a=b(b2+3a)±√a(3b2+a).
In this problem, a=3, b(b2+3a)=b(b2+9) =10, 3b2+a=3b2+3=6, =>b=1

ehild

Last edited: Feb 2, 2012
15. Feb 5, 2012

### S.R

How would you expand this? I'm trying to solve this problem as well.

Last edited: Feb 5, 2012
16. Feb 5, 2012

### Curious3141

Use binomial theorem.

17. Feb 5, 2012

### S.R

I'm having trouble expanding the expression; I'm not sure how to simplify it.

18. Feb 5, 2012

### Curious3141

Post what you've done so far.

19. Feb 5, 2012

### S.R

Would you multiply the expression like this?

$($$\sqrt[3]{10+6\sqrt{3}}$+$\sqrt[3]{10-6\sqrt{3}}$$)$$^3$

Last edited: Feb 5, 2012
20. Feb 5, 2012

### Curious3141

Yes, but you're not multiplying, you're cubing. What's the expansion of the binomial (a+b)3?