Given:

Let

Using partial differentiation:

let

⇒ x² + x + 1 = Ax² + 3Ax + 2A + Bx +2B + Cx² + 2Cx + C

⇒ x² + x + 1 = (2A+2B+C) + (3A+B+2C)x + (A+C)x²

Equating the coefficients of x, x² and constant value. We get:

(a) A + C = 1

(b) 3A + B + 2C = 1

( c) 2A+2B+C =1

After solving we get:

A=-2, B=1 and C=3