\(\frac{a^2}{a^2-b^2-c^2}=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\frac{a^2}{-c\left(a-b\right)-c^2}=\frac{a^2}{-c\left(a-b+c\right)}=\frac{a^2}{2bc}\)
Tương tự \(\Rightarrow P=\frac{a^3+b^3+c^3}{2abc}\)
Mặt khác khi \(a+b+c=0\) dễ dàng chứng minh \(a^3+b^3+c^3=3abc\)
\(\Rightarrow P=\frac{3}{2}\)