Tính $\frac{{\left(a+b+c\right)}^5}{a^5+b^5+c^5}$  thỏa mãn a³ + b³ + c³ = 3abc.

Đặt $A=\frac{{\left(a+b+c\right)}^5}{a^5+b^5+c^5}$ .

Ta có: (a + b + c)³ = a³ + b³ + c³ + 6abc + 3(a²b + ab² + b²c + bc² + c²a + a²c).

⇔ (a + b + c)³ = (3abc + 3a²b + 3ab²) + (3abc + 3b²c + 3bc²) + (3abc + 3c²a + 3a²c).

⇔ (a + b + c)³ = 3ab(a + b + c) + 3bc(a + b + c) + 3ac(a + b + c).

⇔ (a + b + c)³ = 3(ab + bc + ca)(a + b + c).

⇔ (a + b + c).[(a + b + c)² – 3(ab + bc + ca)] = 0.

$\iff \left[\begin{array}{l}a+b+c=0\\ {\left(a+b+c\right)}^2=3\left(ab+bc+ca\right)\end{array}\right.\begin{array}{c}\left(1\right)\\ \left(2\right)\end{array}$

Từ (1), ta có: $A=\frac{{\left(a+b+c\right)}^5}{a^5+b^5+c^5}=\frac{0^5}{a^5+b^5+c^5}=0$ .

Từ (2), ta có: (a + b + c)² = 3(ab + bc + ca).

⇔ a² + b² + c² + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca.

⇔ a² + b² + c² = ab + bc + ca.

⇔ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0.

⇔ (a – b)² + (b – c)² + (c – a)² = 0.

$\iff \left\{\begin{array}{l}a=b\\ b=c\\ c=a\end{array}\right.\iff a=b=c$.

Với a = b = c, ta có: $A=\frac{{\left(a+b+c\right)}^5}{a^5+b^5+c^5}=\frac{{\left(3a\right)}^5}{3a^5}=\frac{3^4.3a^5}{3a^5}=3^4=81$ .

Vậy $A=\left\{\begin{array}{l}0,khia+b+c=0\\ 81,khia=b=c\end{array}\right.$  .