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[kscplay]

Homework Statement

Find the exact value of the following expression.

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

Homework Equations

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.

 

[Curious3141]

Homework Statement

Find the exact value of the following expression.

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

Homework Equations

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.

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

Find $x^3$.

 

[SammyS]

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.

 

[ehild]

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

Find $x^3$.

Ingenious!

ehild

 

[Curious3141]

Ingenious!

ehild

Thanks, but not really!

 

[ehild]

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

ehild

 

[SammyS]

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.

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

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

 

[Curious3141]

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

ehild

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.

 

[kscplay]

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

Find $x^3$.

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.

 

[kscplay]

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

ehild

Could you please elaborate on that? Thanks.

 

[ehild]

Notice that x³=20-6x.

Or choose SammyS's hint...

ehild

 

[ehild]

Could you please elaborate on that? Thanks.

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

 

[kscplay]

Notice that x³=20-6x.

Or choose SammyS's hint...

ehild

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.

 

[ehild]

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.

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=1ehild

 

[S.R]

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

Find $x^3$.

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

 

[Curious3141]

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

Use binomial theorem.

 

[S.R]

Use binomial theorem.

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

 

[Curious3141]

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

Post what you've done so far.

 

[S.R]

Would you multiply the expression like this?

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

 

[Curious3141]

Would you multiply the expression like this?

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

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

 

[S.R]

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

(a+b)^3 = a^3+3a^2b+3ab^2+b^3

Correct?

 

[Curious3141]

(a+b)^3 = a^3+3a^2b+3ab^2+b^3

Correct?

Correct. x^3 = (a+b)^3 = a^3+3a^2b+3ab^2+b^3

So a^3 = ?

b^3 = ?

You can add those up to a simple integer immediately.

That's the easy part, unfortunately.

Express a^2b as (a)(ab) and ab^2 as (b)(ab), so the rest of the expression becomes 3(a+b)(ab).

Can you find an expression for ab?

Use (m+n)(m-n) = m^2 - n^2 for this bit.

Now remember (a+b) = x. You'll now be able to get to a cubic equation in terms of x.

 

[S.R]

Correct. x^3 = (a+b)^3 = a^3+3a^2b+3ab^2+b^3

So a^3 = ?

b^3 = ?

You can add those up to a simple integer immediately.

That's easy part, unfortunately.

Express a^2b as (a)(ab) and ab^2 as (b)(ab), so the rest of the expression becomes 3(a+b)(ab).

Can you find an expression for ab?

Use (m+n)(m-n) = m^2 - n^2 for this bit.

Now remember (a+b) = x. You'll now be able to get to a cubic equation in terms of x.

Will a^3+b^3 = 20?

I'll try the rest tomorrow after school. Thanks for your help.

 

[Curious3141]

Will a^3+b^3 = 20?

I'll try the rest tomorrow after school. Thanks for your help.

Yes, good. OK, till tomorrow then.

 

[SammyS]

Would you multiply the expression like this?

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

Notice that $\displaystyle (10+6\sqrt{3})(10-6\sqrt{3})=(10)^2-(6\sqrt{3})^2=100-108=-8$

This will come in handy when you expand your expression out.

 

[Curious3141]

Notice that $\displaystyle (10+6\sqrt{3})(10-6\sqrt{3})=(10)^2-(6\sqrt{3})^2=100-108=-8$

This will come in handy when you expand your expression out.

Yes, well, we were saving that up for tomorrow.

 

[S.R]

Yes, good. OK, till tomorrow then.

So far I have I calculated a^3+b^3 = 20 and ab = -2. I just need to figure out -6*(a+b). But I'm stuck here.

 

[Mentallic]

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

Let

$$a=\sqrt[3]{10+6\sqrt{3}}$$

$$b=\sqrt[3]{10-6\sqrt{3}}$$

And use the binomial theorem for expanding the cube of a sum

$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

ughh ignore this post, I totally missed page 2 lol

 

[Curious3141]

So far I have I calculated a^3+b^3 = 20 and ab = -2. I just need to figure out -6*(a+b). But I'm stuck here.

That's the neat part! Leave it as -6(a+b) = -6x.

Now put everything together:

x³ = 20 - 6x

x³ + 6x - 20 = 0

Do you know how to solve cubics that have rational roots? Hint: Rational Root Theorem.

 

[S.R]

That's the neat part! Leave it as -6(a+b) = -6x.

Now put everything together:

x³ = 20 - 6x

x³ + 6x - 20 = 0

Do you know how to solve cubics that have rational roots? Hint: Rational Root Theorem.

I didn't notice that; thanks!

EDIT: x^3+6x-20=0

(x-2)(x^2+2x+10) = 0

The only real solution is x=2.

 

[Curious3141]

I didn't notice that; thanks!

x^3-6x-20=0

(x-2)(x^2+2x+10) = 0

The only real solution is x=2.

Yeah, well done!

EDIT: Small correction: the equation should've been x^3 + 6x - 20 = 0. Small matter, just a typo.