How to prove the identity (20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3)=4?
leox
2. Simplify (sqrt(52)-5)^(1/3)* (sqrt(52)+2)^(1/3)
AP
wrote:
>1. (20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3)=4
if x=(20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3)
x^3=40+3(..)^(2/3)(..)(1/3)+3....
x is a root of X^3+pX+q with p=.., q=....
and...
>2. Simplify (sqrt(52)-5)^(1/3)* (sqrt(52)+2)^(1/3)
Ken Pledger
<76fc03df-9bfe-4c81...@k38g2000vbn.googlegroups.com>,
leox <leon...@gmail.com> wrote:
> 1. (20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3)=4
>
> 2. Simplify (sqrt(52)-5)^(1/3)* (sqrt(52)+2)^(1/3)
Is this homework?
The simplest practical way to deal with the first one is to show that
the left-hand side satisfies the cubic equation
x^3 - 6x - 40 = 0,
then check that (x - 4) is the only factor giving a real root. You
can sort out the details.
(However, your expression almost certainly arose by the reverse
process: solving this cubic by Tartaglia's (Cardano's) method. Cardano
himself mentioned that it can be hard to simplify such a sum or
difference of two cube roots.)
Is there a typo in your second problem? Check whether the 2 should
be a 5; in which case you can use (a^k)(b^k) = (ab)^k to simplify the
expression quite a lot.
Ken Pledger.
leox
formulas.
alainv...@gmail.com
> Thanks. But I expected much elementar solution without Cardano's
> formulas.
Bonjour,
Just write:
(2+a*sqrt(2))+(2-a*sqrt(2)) = 4
or
((2+a*sqrt(2))^3)^(1/3) + ((2-a*sqrt(2))^3)^(1/3) = 4
your case corresponds to a=1 .
alain
Mohan Pawar
> Thanks. But I expected much elementar solution without Cardano's
> formulas.
Problem 1:
I think sufficient time has passed for the already sufficiently
disclosed solution. I am leaving only one additional hint here for you
and have the fun of solving it by yourself. Otherwise upon request,
the elementary video solution will be made available on YouTube at
http://www.youtube.com/user/MathsPhysicsClasses
Hints:
(1) Give some serious thought to what Ken Pledger has already
hinted. Forget the detail reference made to "Tartaglia's (Cardano's)
method" but find how he got the
eqn : x^3 - 6x - 40 = 0
(2) assume a = 20 + 40 (2)^1/2 and b = 20 - 40 (2)^1/2
and expand [ (a)^1/3 + (b)^1/3 ]^3
Problem 2:
Yes, there is a strong possibility of typo if the expected answer to
your question is 3
If the question was: Prove that [2 (13)^1/2 - 5 ]^1/3 * [2 (13)^1/2 +
5 ]^1/3 = 3
then use following algebraic identities:
(a)^1/3 * (b)^1/3 = (ab)^1/3
and
(a+b)(a-b) = a^2 - b^2
Best regards.
Mohan Pawar
www.mpClasses.com
-----------------------------------------------------------------------------
Online test preparation in Maths/Physics: ACT/SAT/AP and IIT JEE
-----------------------------------------------------------------------------
US Central Time: 9:08 AM 2/16/2011
alainv...@gmail.com
> On Feb 16, 2:54 am, leox <leonid...@gmail.com> wrote:
>
> > Thanks. But I expected much elementar solution without Cardano's
> > formulas.
>
> Problem 1:
>
> I think sufficient time has passed for the already sufficiently
> disclosed solution. I am leaving only one additional hint here for you
> and have the fun of solving it by yourself. Otherwise upon request,
>
> Hints:
>
> (1) Give some serious thought to what Ken Pledger has already
> hinted. Forget the detail reference made to "Tartaglia's (Cardano's)
> method" but find how he got the
> eqn : x^3 - 6x - 40 = 0
>
> (2) assume a = 20 + 40 (2)^1/2 and b = 20 - 40 (2)^1/2
> and expand [ (a)^1/3 + (b)^1/3 ]^3
>
> Problem 2:
>
> Yes, there is a strong possibility of typo if the expected answer to
> your question is 3
> If the question was: Prove that [2 (13)^1/2 - 5 ]^1/3 * [2 (13)^1/2 +
> 5 ]^1/3 = 3
> then use following algebraic identities:
> (a)^1/3 * (b)^1/3 = (ab)^1/3
> and
> (a+b)(a-b) = a^2 - b^2
>
> Best regards.
>
> Mohan Pawarwww.mpClasses.com
>
> -----------------------------------------------------------------------------
> Online test preparation in Maths/Physics: ACT/SAT/AP and IIT JEE
alainv...@gmail.com
>
> > Thanks. But I expected much elementar solution without Cardano's
> > formulas.
>
> Problem 1:
>
> I think sufficient time has passed for the already sufficiently
> disclosed solution. I am leaving only one additional hint here for you
> and have the fun of solving it by yourself. Otherwise upon request,
>
> Hints:
>
> (1) Give some serious thought to what Ken Pledger has already
> hinted. Forget the detail reference made to "Tartaglia's (Cardano's)
> method" but find how he got the
> eqn : x^3 - 6x - 40 = 0
>
> (2) assume a = 20 + 40 (2)^1/2 and b = 20 - 40 (2)^1/2
> and expand [ (a)^1/3 + (b)^1/3 ]^3
>
> Problem 2:
>
> Yes, there is a strong possibility of typo if the expected answer to
> your question is 3
> If the question was: Prove that [2 (13)^1/2 - 5 ]^1/3 * [2 (13)^1/2 +
> 5 ]^1/3 = 3
> then use following algebraic identities:
> (a)^1/3 * (b)^1/3 = (ab)^1/3
> and
> (a+b)(a-b) = a^2 - b^2
>
> Best regards.
>
> Mohan Pawarwww.mpClasses.com
>
> -----------------------------------------------------------------------------
> Online test preparation in Maths/Physics: ACT/SAT/AP and IIT JEE
> US Central Time: 9:08 AM 2/16/2011
Good afternoon,
Did you read the solution I gave:
((2+a*sqrt(2))^3)^(1/3) + ((2-a*sqrt(2))^3)^(1/3) = 4 ?
Alain
Mohan Pawar
Yes I did. If you had 20 in place of the first 2 in each of the two
terms on LHS, it would be easy for me to know how your hint can be
implemented.
If you meant
((20 +a*sqrt(2))^3)^(1/3) + ((20-a*sqrt(2))^3)^(1/3) = 4 ?
Instead of
((2+a*sqrt(2))^3)^(1/3) + ((2-a*sqrt(2))^3)^(1/3) = 4 ?
then we have the same starting step. But I don't know how your current
first step will lead to the proof I have in mind. But you may also
have a valid solution.
Best regards.
Mohan Pawar
www.mpClasses.com
-----------------------------------------------------------------------------
Online test preparation in Maths/Physics: ACT/SAT/AP and IIT JEE
-----------------------------------------------------------------------------
US Central Time: 9:37 AM 2/16/2011
>
> Alain
alainv...@gmail.com
>
> - Afficher le texte des messages précédents -
Just notice:
for a=1
(2+/-sqrt(2))^3 = 20+/-14sqrt(2)
and (( u)^3)^(1/3) = u
....
A more general solution could be :
((2+b)^3)^(1/3) + ((2-b)^3)^(1/3) = 4 ,
with b anything you want 7^(1/5) , sqrt(2)+sqrt(3)..
Alain
Mohan Pawar
> On Feb 16, 2:54 am, leox <leonid...@gmail.com> wrote:
>
> > Thanks. But I expected much elementar solution without Cardano's
> > formulas.
>
> Problem 1:
>
> I think sufficient time has passed for the already sufficiently
> disclosed solution. I am leaving only one additional hint here for you
> and have the fun of solving it by yourself. Otherwise upon request,
>
> Hints:
>
> (1) Give some serious thought to what Ken Pledger has already
> hinted. Forget the detail reference made to "Tartaglia's (Cardano's)
> method" but find how he got the
> eqn : x^3 - 6x - 40 = 0
>
> (2) assume a = 20 + 40 (2)^1/2 and b = 20 - 40 (2)^1/2
An obvious typo in the line above. 40 is not in the problem but 14 is.
Corrected line: assume a = 20 + 14(2)^1/2 and b = 20 - 14(2)^1/2
--
Mohan Pawar
US Central Time: 1:57 PM 2/16/2011
Ray Vickson
> 1. (20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3)=4
If you have access to Maple, you can find the minimial polynomial of a
= (20+14*sqrt(2))^(1/3)+ (20-14*sqrt(2))^(1/3). In this case, Maple
9.5 gives:
with(PolynomialTools);
MinimalPolynomial(a,3);
-4 + _X
In other words, the polynomial m(x) of smallest degree (degree <= 3)
that has 'a' as a root is f(x) = x-4. Of course, that means that a =
4. Maple applies some standard algorithms to get m(x), so one can
apply those manually to get a proof.
R.G. Vickson
>
> 2. Simplify (sqrt(52)-5)^(1/3)* (sqrt(52)+2)^(1/3)