75. Let positive real numbers $a, b, c$ satisfy the condition $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13$. Find the maximum and minimum values of the following expression.
$$P=\frac{a^{3}+b^{3}+c^{3}}{a b c} \text { (Pham Kim Hung) }$$
Let $x=\sum_{c y c} \frac{a}{b} ; y=\sum_{c y c} \frac{b}{a} ; m=\sum_{c y c} \frac{a^{2}}{b c} ; n=\sum_{c y c} \frac{b c}{a^{2}}$; then we have $x+y=10$ and
$$x^{3}=3(m+n)+6+\sum_{c y c} \frac{a^{3}}{b^{3}} ; \quad y^{3}=3(m+n)+6+\sum_{c y c} \frac{b^{3}}{a^{3}}$$

From the above identities, we have
$$x^{3}+y^{3}=\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)+6(m+n)+9$$

We also have
$$\begin{array}{l}
m n=3+\sum_{c y y} \frac{a^{3}}{b^{3}}+\sum_{c y c} \frac{b^{3}}{a^{3}}=\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right) \text { and } \\
x y=\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=3+m+n
\end{array}$$

Thus, we have
$$10^{3}-30(3+m+n)=m n+6(m+n)+9 \Leftrightarrow 10^{3}-99=m n+36(m+n)$$

So $m, n$ are the two positive roots of the following quadratic equation
$$f(t)=t^{2}-(x y-3) t+(1009-36 x y)$$

Let $r=x y$, we can determine
$$2 m=(x y-3) \pm \sqrt{(x y-3)^{2}-4(1009-36 x y)}$$

Consider the function $g(r)=r-3-\sqrt{r^{2}+138 r-4027} \quad(0 \leq r \leq 25)$. Notice that
$$g^{\prime}(r)=1-\frac{2 r+138}{2 \sqrt{r^{2}+138 r-4027}}<0$$

Therefore, we have $11-2 \sqrt{3} \leq m \leq 11+2 \sqrt{3}$, with equality holding when $x-y=(a-b)(b-c)(c-a)=0$. The minimum value of $m$ is $11-2 \sqrt{3}$, achieved when $a=b=(2+\sqrt{3}) c$ and its permutations. The maximum value of $m$ is $11+2 \sqrt{3}$, achieved when $a=b=(2-\sqrt{3}) c$ and its permutations.