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- Feb 14, 2012

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- Thread starter
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- #1

- Feb 14, 2012

- 3,680

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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Congratulations to the following members for their correct solutions:

1. MarkFL

2. kaliprasad

3. soroban

4. mente oscura

Solution from kaliprasad:

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}=\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:

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