How Do You Solve the System Involving $a$ and $b$ in POTW #202?

[anemone]

Here is this week's POTW:

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Find all real numbers for $a$ and $b$ that satisfy the system of equations below:

\(\displaystyle \frac{1}{a}+\frac{1}{2b}=(a^2+3b^2)(3a^2+b^2)\)

\(\displaystyle \frac{1}{a}-\frac{1}{2b}=2(b^4-a^4)\)

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem. :(

You can find the proposed solution below:

Spoiler

We're given the system

\(\displaystyle \frac{1}{a}+\frac{1}{2b}=(a^2+3b^2)(3a^2+b^2)\)

\(\displaystyle \frac{1}{a}-\frac{1}{2b}=2(b^4-a^4)\)

Adding the equations gives

\(\displaystyle \frac{2}{a}=a^4+10a^2b^2+5b^4\)---(1)

Subtracting the equations gives

\(\displaystyle \frac{1}{b}=5a^4+10a^2b^2+b^4\)---(2)

Now, multiply the equation (1) by $a$ and get:

\(\displaystyle 2=a^5+10a^3b^2+5ab^4\)---(3)

Next, we multiply the equation (2) by $b$ and get:

\(\displaystyle 1=5a^4b+10a^2b^3+b^5\)---(4)

Adding and subtracting the equations (3) and (4) yield

\(\displaystyle 3=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=(a+b)^5\)

\(\displaystyle 1=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5=(a-b)^5\)

Solving both last two equations for $a$ and $b$ we get

\(\displaystyle \left(a,\,b\right)=\left(\frac{1+\sqrt[5]{3}}{2},\,\frac{\sqrt[5]{3}-1}{2}\right)\)