(x+3)(x2−x+1)x2+x+1=x+3A+x2−x+1Bx+C
x2+x+1=A(x2−x+1)+(Bx+C)(x+3)
Let's find A, B and C by equating corresponding coefficients:
1=A+B (1)
1=−A+3B+C (2)
1=A+3C (3)
Substracting (3) from (1) we obtain:
B=3C (4)
Substitution (4) into (2) gives
1=−A+10C (5)
Adding (3) and (5) we obtain
C=132
B=136
A=137
Answer:
(x+3)(x2−x+1)x2+x+1=137x+31+132x2−x+13x+1
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