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Problem of the week #102 - March 10th, 2014

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anemone

MHB POTW Director
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Feb 14, 2012
3,709
Represent the number $\sqrt[3]{1342\sqrt{167}+2005}$ in the form where it contains only addition, subtraction, multiplication, division and square roots.

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anemone

MHB POTW Director
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Feb 14, 2012
3,709
Re: Problem of the week #102 -March 10th, 2014

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. kaliprasad
3. soroban
4. mente oscura

Solution from kaliprasad:

Because 167 is prime so the cube root shall be of the form $a + b\sqrt{167}$

Cube both sides and equate to cube of RHS to get

$a^3 + 3a^2b\sqrt {167} + 3a * 167 b^2 + 167 b^3 \sqrt {167} = 2005 + 1342 \sqrt {167}$

Equate the rational parts on both sides to get

$a^3 + 501 ab^2 = 2005$ ...(1)

Equate their rational parts on both sides to get

$3a^2 b + 167 b^3 = 1342$ ... (1)

First check for positive integers and we see that a = 1 and b =2 satisfy (1) and (2) and hence solution is

$1 + 2 \sqrt {167}$


Solution from mente oscura:

[tex]\sqrt[3]{1342 \sqrt{167}+2005}[/tex]


[tex]\sqrt[3]{1342 \sqrt{167}+2005}=\sqrt[3]{(a+b)^3}[/tex]

[tex]a^3+3a^2b+3ab^2+b^3=1342 \sqrt{167}+2005[/tex]

[tex]Let \ a=x \sqrt{167}[/tex]

Therefore:

[tex]a^3+3ab^2=1342 \sqrt{167} \rightarrow{} 167x^3 \cancel{\sqrt{167}}+3x \cancel{\sqrt{167}} \ b^2=1342 \cancel{\sqrt{167}}[/tex]

[tex]b=\sqrt{\dfrac{1342-167x^3}{3x}}[/tex]

[tex]For \ x=2 \rightarrow{}b=1 \ and \ a=2 \sqrt{167}[/tex]

Same:

[tex]3x^2 \ 167b+b^3=2005[/tex]

[tex]For \ b=1 \rightarrow{} x=2[/tex]

Solution:
[tex]\sqrt[3]{1342 \sqrt{167}+2005}=2 \sqrt{167}+1[/tex]
 
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