[abhishek sinha]

Note \(2a^2+bc= a^2-a(b+c)+bc= -(a-b)(c-a)\). Similarly, \(2b^2+ca=-(a-b)(b-c)\) and \(2c^2+ab=-(b-c)(c-a)\). Hence,

\[ P= - \frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}=1.\]

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