For real numbers $x$ and $y$, where $x \neq y$, simplify the expression

$$
\left(\left(\frac{2 x+y}{x-y}\right)^{3}+1\right) \cdot(x-y)^{3}:\left(9 x^{3}-9 y^{3}\right)
$$

and write the result as a simplified fraction.

(10 points)
1. We cube the fraction by cubing the numerator separately \((2 x+y)^{3}=8 x^{3}+12 x^{2} y+6 x y^{2}+y^{3}\) and the denominator separately \((x-y)^{3}=x^{3}-3 x^{2} y+3 x y^{2}-y^{3}\). We find the common denominator of the expression in the parentheses (e.g., \(x^{3}-3 x^{2} y+3 x y^{2}-y^{3}\)), so we extend the number 1 to the common denominator, add the fractions, and simplify the resulting fraction \(\left(\frac{2 x+y}{x-y}\right)^{3}+1=\frac{9 x^{3}+9 x^{2} y+9 x y^{2}}{(x-y)^{3}}\).

We multiply the resulting fraction by \((x-y)^{3}\) and get \(9 x^{3}+9 x^{2} y+9 x y^{2}\). We write the division in the form of a fraction and factor out the common factor in the numerator \(9 x\left(x^{2}+x y+y^{2}\right)\) and factor the denominator \(9(x-y)\left(x^{2}+x y+y^{2}\right)\). We simplify the fraction and get the result \(\frac{x}{x-y}\).

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-11.jpg?height=565&width=1759&top_left_y=757&top_left_x=157)