6.007. $\frac{x^{2}}{a^{3}}+\frac{b^{3}}{x^{2}}=\frac{b}{a}+\frac{b^{2}}{a^{2}}$.
## Solution.

Domain of definition: $x \neq 0, a \neq 0$.

$$
\frac{x^{2}}{a^{3}}+\frac{b^{3}}{x^{2}}=\frac{b}{a}+\frac{b^{2}}{a^{2}} \Leftrightarrow \frac{x^{4}-\left(a^{2} b+a b^{2}\right) x^{2}+a^{3} b^{3}}{a^{3} x^{2}}=0
$$

or $x^{4}-\left(a^{2} b+a b^{2}\right) x^{2}+a^{3} b^{3}=0, a x \neq 0$.

The equation is a biquadratic equation in terms of $x$. Let $x^{2} = y$, then our equation becomes

$$
y^{2}-\left(a^{2} b+a b^{2}\right) y+a^{3} b^{3}=0
$$

from which

$$
\begin{aligned}
& y_{1,2}=\frac{a^{2} b+a b^{2} \pm \sqrt{\left(a^{2} b+a b^{2}\right)^{2}-4 a^{3} b^{3}}}{2}= \\
& =\frac{a^{2} b+a b^{2} \pm \sqrt{a^{4} b^{2}+2 a^{3} b^{3}+a^{2} b^{4}-4 a^{3} b^{3}}}{2}= \\
& =\frac{a^{2} b+a b^{2} \pm \sqrt{a^{4} b^{2}-2 a^{3} b^{3}+a^{2} b^{4}}}{2}=\frac{a^{2} b+a b^{2} \pm \sqrt{\left(a^{2} b-a b^{2}\right)^{2}}}{2}= \\
& =\frac{a^{2} b+a b^{2} \pm\left(a^{2} b-a b^{2}\right)}{2} ; y_{1}=a b^{2}, y_{2}=a^{2} b
\end{aligned}
$$

To find $x$, we need to solve two quadratic equations:

$x^{2}=a b^{2}$ or $x^{2}=a^{2} b$.

$x_{1,2}= \pm \sqrt{a b^{2}}= \pm b \sqrt{a}$, where considering the domain of definition $a>0$;

$x_{3,4}= \pm \sqrt{a^{2} b}= \pm a \sqrt{b}$, where $b \geq 0$.

Answer: $x_{1,2}= \pm b \sqrt{a}$, where $a>0 ; x_{3,4}= \pm a \sqrt{b}$, where $b \geq 0$.