a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Suy ra:

 \(\frac{a}{b+c}=\frac{1}{2}\Rightarrow a=\frac{b+c}{2}=\frac{1}{2}\times\left(b+c\right)\)

\(\frac{b}{a+c}=\frac{1}{2}\Rightarrow b=\frac{a+c}{2}=\frac{1}{2}\times\left(a+c\right)\)

\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow c=\frac{a+b}{2}=\frac{1}{2}\times\left(a+b\right)\)

Thay  \(a=\frac{1}{2}\times\left(b+c\right)\);  \(b=\frac{1}{2}\times\left(a+c\right)\); \(c=\frac{1}{2}\times\left(a+b\right)\) vào P ta được:

\(\frac{b+c}{\frac{1}{2}\times\left(b+c\right)}+\frac{c+a}{\frac{1}{2}\times\left(a+c\right)}+\frac{a+b}{\frac{1}{2}\times\left(a+b\right)}\)

\(=\frac{\text{ }1\text{ }}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}\)

\(=2+2+2=6\)

Vậy giá trị của P  là 6

Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

Ta có \(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\)

\(=1+\frac{c}{a-b}.\frac{b^2-bc+ca-a^2}{ab}\)

\(=1+\frac{c}{a-b}.\frac{\left(b-a\right)\left(a+b-c\right)}{ab}=1+\frac{2c^2}{ab}\)

Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^2}{bc};M.\frac{b}{c-a}=1+\frac{2b^2}{ca}\)

Do vậy \(A=3+2.\frac{a^3+b^3+c^3}{abc}=9\left(do.a+b+c=0.thi.a^3+b^3+c^3=3abc\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(...=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)( a, b, c khác 0 )

=> \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=1\)

=> \(\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Thế vào P ta được :

\(P=\frac{2c}{a}\cdot\frac{2a}{b}\cdot\frac{2b}{c}=\frac{8abc}{abc}=8\)