f(x)=1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x1+(a2+b2+c2+2)x(1+b2)x1+b2x(1+b2)x(1+c2)x(1+c2)x1+c2x, Applying C1→C1+C2+C3=111(1+b2)x1+b2x(1+b2)x(1+c2)x(1+c2)x1+c2x∵a2+b2+c2+2=0f(x)=001x−11−x(1+b2)x0x−11+cx; Applying R1→R1−R2,R2→R2−R3f(x)=(x−1)2 Hence degree =2.