Ta có:$(a+b+c)^{2}=2(a^{2}+b^{2}+c^{2})\Leftrightarrow a^{2}+b^{2}+c^{2}=2(ab+bc+ca)$
$\Leftrightarrow (a+b+c)^{2}=4(ab+bc+ca)$
$\Rightarrow P=\frac{4(a^{3}+b^{3}+c^{3})}{(a+b+c)^{3}}=4(\Sigma (\frac{a}{a+b+c})^{3})$
Đặt $x=\frac{a}{a+b+c};y=\frac{b}{a+b+c};z=\frac{c}{a+b+c}$
$\Rightarrow x+y+z=1$&$xy+yz+zx=\frac{ab+bc+ca}{(a+b+c)^{2}}=\frac{1}{4}$
$\Rightarrow \begin{cases}y+z=1-x\\ yz+x(y+z)=\frac{1}{4} \end{cases}\Leftrightarrow\begin{cases}y+z=1-x \\ yz=x^{2}-x+\frac{1}{4}\end{cases}$
Mà:$(y+z)^{2}\geq 4yz\Rightarrow 0\leq x\leq\frac{2}{3}$
Khi đó: $P=4(x^{3}+y^{3}+z^{3})=4\left[ {x^{3}+(y+z)^{3}-3yz(y+z)} \right]=12x^{3}-12x^{2}+3x+1=f(x)$
Xét h/s$f(x) $ liên tục trên $\left[ {0;\frac{2}{3}} \right]$
$f'(x)=36x^{2}-24x+3;f'(x)=0\Leftrightarrow x=\frac{1}{6} or x=\frac{1}{2}$
$\Rightarrow 1\leq P\leq \frac{11}{9}$
Min P=1 đạt tại $a=0;b=c$
Max$P=\frac{11}{9} $đạt tại $b=c=\frac{a}{2} or b=c=\frac{5a}{2}$