MHB What are the real pairs $(a,\,b)$ that satisfy this system of equations?

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The discussion focuses on finding real pairs (a, b) that satisfy the equations 4(b² - a²) = 3√[3]{a²b⁵} and b² + a² = 5√[3]{a⁴b}. Participants engage in sharing partial solutions and express gratitude for contributions. The conversation includes light-hearted remarks, indicating a friendly atmosphere. However, no definitive solutions or conclusions are reached within the thread. The exploration of the equations remains ongoing.
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Determine all real paris $(a,\,b)$ that satisfy the system of equations below:

$4(b^2-a^2)=3\sqrt[3]{a^2b^5}$

$b^2+a^2=5\sqrt[3]{a^4b}$
 
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anemone said:
Determine all real paris $(a,\,b)$ that satisfy the system of equations below:

$4(b^2-a^2)=3\sqrt[3]{a^2b^5}$

$b^2+a^2=5\sqrt[3]{a^4b}$

Hello.
4(b^2-a^2)(b^2+a^2)=3\sqrt[3]{a^2b^5} \ 5\sqrt[3]{a^4b}

4(b^4-a^4)=15a^2b^2

4a^4+15a^2b^2-4b^4=0

a^2=\dfrac{-15b^2 \pm \sqrt{289b^4}}{8}

Real solution:

a^2=\dfrac{b^2}{4} \rightarrow{}a=\pm \dfrac{b}{2}

Regards.
 
mente oscura said:
Hello.
4(b^2-a^2)(b^2+a^2)=3\sqrt[3]{a^2b^5} \ 5\sqrt[3]{a^4b}

4(b^4-a^4)=15a^2b^2

4a^4+15a^2b^2-4b^4=0

a^2=\dfrac{-15b^2 \pm \sqrt{289b^4}}{8}

Real solution:

a^2=\dfrac{b^2}{4} \rightarrow{}a=\pm \dfrac{b}{2}

Regards.

That is a good "partial" solution, mente oscura! :pHehehe...and thanks for participating!
 
anemone said:
That is a good "partial" solution, mente oscura! :pHehehe...and thanks for participating!

Hello.

Yes. It is that I am a bit lazy.

4(b^2-\dfrac{b^2}{4})=3 \sqrt[3]{\dfrac{b^7}{4}}

Solutions: (0,0), (2,4), (-2,4)

Regards.
 
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