Evaluate expression W/O Calculator

I multiplied the expression out, but I still don't see how to get the a^3 from the b^3, and vice versa. \left(x^3+6(x-2)\right)^3=x^9+18x^6+108x^3+216x^2-864x+864=0\noindent \left(x^3+6(x-2)\right)^3=x^9+18x^6+108x^3+216x^2-864x+864=0You mean (x3-8+6x)3? You have to expand the first term of the expansion too. \left(x^3+6(x-2)\
  • #1
kscplay
23
0

Homework Statement



Find the exact value of the following expression.

[itex]\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}[/itex]

Homework Equations



The Attempt at a Solution



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

Homework Statement



Find the exact value of the following expression.

[itex]\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}[/itex]

Homework Equations



The Attempt at a Solution



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

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

Find [itex]x^3[/itex].
 
  • #3
Maybe there are values for a & b such that
[itex]\displaystyle \left(a+b\sqrt{3}\right)^3=10+6\sqrt{3}[/itex]​
and
[itex]\displaystyle \left(a-b\sqrt{3}\right)^3=10-6\sqrt{3}[/itex]​
...

or something like that.
 
  • #4
Curious3141 said:
Let [itex]x = \sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}[/itex]

Find [itex]x^3[/itex].

Ingenious!:cool:

ehild
 
  • #5
ehild said:
Ingenious!:cool:

ehild

Thanks, but not really! :biggrin:
 
  • #6
Why not? You get an easily factorizable equation of form x^3+ax+b=0.

ehild
 
  • #7
SammyS said:
Maybe there are values for a & b such that
[itex]\displaystyle \left(a+b\sqrt{3}\right)^3=10+6\sqrt{3}[/itex]​
and
[itex]\displaystyle \left(a-b\sqrt{3}\right)^3=10-6\sqrt{3}[/itex]​
...

or something like that.
The answer is: a=1, b=1 .

[itex]\displaystyle \left(1\pm\sqrt{3}\right)^3=10\pm6\sqrt{3}[/itex]
 
  • #8
ehild said:
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. :smile:

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
Curious3141 said:
Let [itex]x = \sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}[/itex]

Find [itex]x^3[/itex].

Okay, I get the following for x3:

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


I still don't see it. Is my x3 correct? Thanks.
 
  • #10
ehild said:
Why not? You get an easily factorizable equation of form x^3+ax+b=0.

ehild

Could you please elaborate on that? Thanks.
 
  • #11
Notice that x3=20-6x.

Or choose SammyS's hint...

ehild
 
  • #12
kscplay said:
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: [tex]\displaystyle \left(1\pm\sqrt{3}\right)^3=10\pm6\sqrt{3}[/tex]
He is the Master of Square Roots :cool:

ehild
 
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  • #13
ehild said:
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.
 
  • #14
kscplay said:
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. :smile: 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
 
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  • #15
Curious3141 said:
Let [itex]x = \sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}[/itex]

Find [itex]x^3[/itex].

How would you expand this? I'm trying to solve this problem as well.
 
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  • #16
S.R said:
How would you expand this? I'm trying to solve this problem as well.

Use binomial theorem.
 
  • #17
Curious3141 said:
Use binomial theorem.

I'm having trouble expanding the expression; I'm not sure how to simplify it.
 
  • #18
S.R said:
I'm having trouble expanding the expression; I'm not sure how to simplify it.

Post what you've done so far.
 
  • #19
Would you multiply the expression like this?

[itex]([/itex][itex]\sqrt[3]{10+6\sqrt{3}}[/itex]+[itex]\sqrt[3]{10-6\sqrt{3}}[/itex][itex])[/itex][itex]^3[/itex]
 
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  • #20
S.R said:
Would you multiply the expression like this?

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

Yes, but you're not multiplying, you're cubing. What's the expansion of the binomial (a+b)3?
 
  • #21
Curious3141 said:
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?
 
  • #22
S.R said:
(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.
 
  • #23
Curious3141 said:
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.
 
  • #24
S.R said:
Will a^3+b^3 = 20?

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

Yes, good. OK, till tomorrow then.
 
  • #25
S.R said:
Would you multiply the expression like this?

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

This will come in handy when you expand your expression out.
 
  • #26
SammyS said:
Notice that [itex]\displaystyle (10+6\sqrt{3})(10-6\sqrt{3})=(10)^2-(6\sqrt{3})^2=100-108=-8[/itex]

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

Yes, well, we were saving that up for tomorrow. :biggrin:
 
  • #27
Curious3141 said:
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.
 
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  • #28
S.R said:
How would you expand this? I'm trying to solve this problem as well.

Let

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

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

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

[tex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex]

ughh ignore this post, I totally missed page 2 lol
 
  • #29
S.R said:
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.
 
  • #30
Curious3141 said:
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.
 
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  • #31
S.R said:
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! :approve:

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

1. How do I evaluate an expression without a calculator?

To evaluate an expression without a calculator, you can use the order of operations (PEMDAS) to simplify the expression. Start by solving any operations within parentheses, then move on to exponents, multiplication and division (from left to right), and finally addition and subtraction (from left to right).

2. What is the order of operations?

The order of operations, also known as PEMDAS, stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This is the order in which you should solve an expression without a calculator.

3. Can I use a calculator to check my answer after evaluating an expression?

Yes, you can use a calculator to check your answer after evaluating an expression. However, it is important to try to solve the expression without a calculator first to improve your math skills and understanding of the concept.

4. What if the expression contains variables?

If the expression contains variables, you can still use the order of operations to simplify it. Simply substitute the variables with their respective values and follow the order of operations to evaluate the expression.

5. Are there any shortcuts or tricks to evaluate an expression without a calculator?

There are a few shortcuts or tricks that can help you evaluate an expression without a calculator. For example, you can use the commutative and associative properties to rearrange the order of operations and make the expression easier to solve. You can also use mental math strategies, such as rounding or breaking down numbers, to simplify the calculations.

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