Cubic equation

[sakibtanvir]

IF\[8ab(a^2+b^2)=5\]\[a,b\in \mathbb{R}\]Find the values of $a$ and $b$ .

[*Mahi*]

restrictions for $a,b$? (Like $a,b \in \mathbb N$)

[sakibtanvir]

It is edited...

[nafistiham]

$(a,b)=(\frac{\sqrt3+\sqrt2}{2},\frac{\sqrt3-\sqrt2}{2})$

*এইটার মান দিয়ে তো প্রবেশিকায় বারবার আসে ।

[Phlembac Adib Hasan]

IF\[8ab(a^2+b^2)=5\]\[a,b\in \mathbb{R}\]Find the values of $a$ and $b$ .

Infinite solutions.My calculation shows that there is at least one $b$ for every $a\in \mathbb{R}-0$.For example, $a=1,b=\frac {1}{2}$ satisfies the equation.

Let $b=x$ is a variable.Then the equation becomes \[8ax^3+8a^3x=5\]\[\Rightarrow x^3+a^2x=\frac {5}{8a}\]\[\Rightarrow x_1=\sqrt[3]{\frac{5}{16a}+\sqrt{\frac{25}{256a^2}+\frac{a^6}{27}}}-\sqrt[3]{-\frac{5}{16a}+\sqrt{\frac{25}{256a^2}+\frac{a^6}{27}}}\]
As $\frac{25}{256a^2}+\frac{a^6}{27}$ is positive for every real $a$ except $0$, it implies at least one of the roots of the first cubic equation is real.So $8ab(a^2+b^2)$ is satisfied by every real number $a$ except $0$.

[sakibtanvir]

$(a,b)=(\frac{\sqrt3+\sqrt2}{2},\frac{\sqrt3-\sqrt2}{2})$

*এইটার মান দিয়ে তো প্রবেশিকায় বারবার আসে ।

I agree.A very familiar problem...But it is not that easy when we do in the reversed way...
By the way,I am editing this...Prove that,\[a=\frac{\sqrt{3}+\sqrt{2}}{2},b=\frac{\sqrt{3}-\sqrt{2}}{2}\]