$\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}$

Suy ra:

*2.(a + b) = 3(2a – b)

2a +  2b = 6a – 3b

4a = 5b $\Rightarrow $ $a=\frac{5}{4}b$ (1)

*3(b – c + a)=2(2a – b)

3b – 3c + 3a = 4a – 2b

a = 5b – 3c (2)

Từ (1) và (2) ta có:

$a=\frac{5}{4}b$= 5b – 3c $\Rightarrow \frac{15}{4}b=3c\Rightarrow c=\frac{5}{4}b$ (3)

Thay (1) và (3) vào P ta được:

P=$\frac{{{(5b+4a)}^{5}}}{{{(5b+4c)}^{2}}{{(a+3c)}^{3}}}$=$\frac{{{\left( 5b+4.\frac{5}{4}b \right)}^{5}}}{{{\left( 5b+4.\frac{5}{4}b \right)}^{2}}{{\left( \frac{5}{4}b+\frac{15}{4}b \right)}^{3}}}$=$\frac{{{\left( 5b+4.\frac{5}{4}b \right)}^{5}}}{{{\left( 5b+4.\frac{5}{4}b \right)}^{2}}{{\left( \frac{5}{4}b+\frac{15}{4}b \right)}^{3}}}$

=$\frac{{{\left( 10b \right)}^{5}}}{{{\left( 10b \right)}^{2}}{{\left( 5b \right)}^{3}}}=8$