Let $x,y,z>0$ such that
$$(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)=\frac{91}{10}$$
Compute
$$\left[(x^3+y^3+z^3)\left(\frac1{x^3}+\frac1{y^3}+\frac1{z^3}\right)\right]$$
where $[.]$ represents the integer part.

[i]Proposed by Marian Cucoanoeş and Marius Drăgan[/i]
1. Given the condition:
   \[
   (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = \frac{91}{10}
   \]
   We need to compute:
   \[
   \left\lfloor (x^3 + y^3 + z^3) \left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right) \right\rfloor
   \]
   where $\lfloor \cdot \rfloor$ denotes the integer part.

2. Without loss of generality (WLOG), assume \( x \leq y \leq z \). This assumption helps in simplifying the problem by providing a structure to the inequalities.

3. From the given condition, we can derive:
   \[
   (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = \frac{91}{10}
   \]
   Expanding the left-hand side, we get:
   \[
   (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = 3 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + \frac{y}{z} + \frac{z}{x} + \frac{z}{y}
   \]
   Therefore:
   \[
   3 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + \frac{y}{z} + \frac{z}{x} + \frac{z}{y} = \frac{91}{10}
   \]
   Simplifying, we get:
   \[
   \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + \frac{y}{z} + \frac{z}{x} + \frac{z}{y} = \frac{61}{10}
   \]

4. To estimate the bounds, we use the assumption \( x \leq y \leq z \). This implies:
   \[
   \frac{z}{x} \leq \frac{5}{4}
   \]
   and thus:
   \[
   \left(\frac{z}{x}\right)^3 \leq \left(\frac{5}{4}\right)^3 = \frac{125}{64}
   \]

5. Now, we need to compute:
   \[
   (x^3 + y^3 + z^3)\left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right)
   \]
   Using the bounds derived, we estimate:
   \[
   (x^3 + y^3 + z^3) \leq 2\left(\frac{5}{4}\right)^3 + 1^3 = 2 \cdot \frac{125}{64} + 1 = \frac{250}{64} + 1 = \frac{250}{64} + \frac{64}{64} = \frac{314}{64} = 4.90625
   \]
   and:
   \[
   \left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right) \leq 2\left(\frac{4}{5}\right)^3 + 1^3 = 2 \cdot \frac{64}{125} + 1 = \frac{128}{125} + 1 = \frac{128}{125} + \frac{125}{125} = \frac{253}{125} = 2.024
   \]

6. Therefore:
   \[
   (x^3 + y^3 + z^3)\left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right) \leq 4.90625 \cdot 2.024 = 9.93025
   \]
   and:
   \[
   (x^3 + y^3 + z^3)\left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right) \geq 9
   \]

7. Since the value is between 9 and 10, the integer part is:
   \[
   \left\lfloor (x^3 + y^3 + z^3) \left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right) \right\rfloor = 9
   \]

The final answer is \(\boxed{9}\)